If the 2000 numbers are of the form $10^n$, for n from 1 to 2000, this seems to be wrong.

We'll assume otherwise and argue by contradiction. Denote the set by $\{a_1, a_2, \dots, a_{2000}\}$, with $a_1 < \dots, < a_{2000}$ and set $d_i = a_{i + 1} - a_i > 0$. There are in all $\binom{2000}{2} = 1999000$ positive differences of the form $a_i - a_j$ ($i > j$). When we order these differences, each must be at least $10^5/(10^5 - 1)$ times the previous lest the problem condition would hold. Thus the largest of these differences must be at least \[\left(\frac{10^5}{10^5 - 1}\right)^{1999000} > 4 \times 10^8 \] times as large as the smallest. The largest possible difference is of course \[a_{2000} - a_1 = d_1 + d_2 + \dots + d_{1999} \le 1999\max{d_k}\] Hence we need \[1999 \left(\frac{\max{d_k}}{\min{d_k}}\right) > 4 \times 10^{8} \implies \frac{\max{d_k}}{\min{d_k}} > 2 \times 10^5.\] Lemma : $\max{d_k} \le 10^5\min{d_k}$. Proof : Suppose $d_{k_1} = \max{d_k}$ and $d_{k_2} = \min{d_k}$. If the lemma were not to hold, and $k_1 > k_2$, we have \[1 < \frac{d_{k_2} + \dots d_{k_1}}{d_{k_2 + 1} + \dots d_{k_1}} < \frac{10^5 + 1}{10^5}.\] If instead $k_1 < k_2$, \[\frac{10^5 - 1}{10^5} < \frac{d_{k_1} + \dots + d_{k_2 - 1}}{d_{k_1} + \dots + d_{k_2}} < 1. \; \; \; \; \; \; \; \; \; \; \blacksquare\] The conclusion follows.

SCP wrote: If the 2000 numbers are of the form $10^n$, for n from 1 to 2000, this seems to be wrong. Wrong: $| \frac{10^{2000}-10^1}{10^{2000}-10^2}-1|=| \frac{9}{10^{1999}-1}|< \frac{1}{100000}$

Can we find the best constant to replace $\frac1{100000}$?

^ At MOP we determined that it's somewhere around 500 million, but I might be mis-remembering.

$2000$ and $\frac{1}{10000}$ can be replaced with any $n \in N$ and $r \in R$ such that \[ r (1+r)^{\frac{n(n+1)}{2}-1} > 2.\]

Some of us in black MOP initially had much trouble lower bounding $\left(\frac{10^5}{10^5 - 1}\right)^{1999000}$. We managed to come up with the bound \[\left(\frac{10^5}{10^5 - 1}\right)^{1999000}>1+1999000\cdot\frac{1}{99999}>20\] I, for instance, first found a binomial theorem bash proving that the $\dbinom{1999000}{9}$ term was big enough before realizing we could do $(1+\frac{1}{\epsilon})^{k\epsilon}>(1+\epsilon\frac{1}{\epsilon})^k=2^k$ rather than just $> 1 + k$

Davi Medeiros wrote: SCP wrote: If the 2000 numbers are of the form $10^n$, for n from 1 to 2000, this seems to be wrong. Wrong: $| \frac{10^{2000}-10^1}{10^{2000}-10^2}-1|=| \frac{9}{10^{1999}-1}|< \frac{1}{100000}$ Yeah, but you use $a=10^{2000}=c$, which is wrong because you shoud have $a\ne c$.

lyukhson wrote: ... $a \neq c$ or $b \neq d $ ... Read the statement again ... it is enough to have $b\neq d$.

Oh, yeah, I cannot read ... thanks

Dear hyperbolictangent , about your solution , you said that : When we order these differences, each must be at least $ frac{ 10^5}{10^5 - 1} $ times the previous lest the problem condition would hold. I think that this sentence isn't quite precise, because what if after ordering we have, for example: $a_3 -a_4 $ and then $ a_3 - a_5 $ then u cant say that the first one is at least $ frac{ 10^5}{10^5 - 1} $ times the previous. Am i mistaken? please someone reply

Note that the problem statement only requires that either $a \ne c$ or $b \ne d$, and not that both hold. Thus we can legally set $a = a_3$, $b = a_4$, $c = a_3$, and $d = a_5$.

For $n$ numbers the best constant on the right hand side is $\frac{2}{n(n+1)}$.

LevonNurbekian wrote: For $n$ numbers the best constant on the right hand side is $\frac{2}{n(n+1)}$. Nope

hyperbolictangent wrote: We'll assume otherwise and argue by contradiction. Denote the set by $\{a_1, a_2, \dots, a_{2000}\}$, with $a_1 < \dots, < a_{2000}$ and set $d_i = a_{i + 1} - a_i > 0$. There are in all $\binom{2000}{2} = 1999000$ positive differences of the form $a_i - a_j$ ($i > j$). When we order these differences, each must be at least $10^5/(10^5 - 1)$ times the previous lest the problem condition would hold. Thus the largest of these differences must be at least \[\left(\frac{10^5}{10^5 - 1}\right)^{1999000} > 4 \times 10^8 \]times as large as the smallest. The largest possible difference is of course \[a_{2000} - a_1 = d_1 + d_2 + \dots + d_{1999} \le 1999\max{d_k}\]Hence we need \[1999 \left(\frac{\max{d_k}}{\min{d_k}}\right) > 4 \times 10^{8} \implies \frac{\max{d_k}}{\min{d_k}} > 2 \times 10^5.\] Lemma : $\max{d_k} \le 10^5\min{d_k}$. Proof : Suppose $d_{k_1} = \max{d_k}$ and $d_{k_2} = \min{d_k}$. If the lemma were not to hold, and $k_1 > k_2$, we have \[1 < \frac{d_{k_2} + \dots d_{k_1}}{d_{k_2 + 1} + \dots d_{k_1}} < \frac{10^5 + 1}{10^5}.\]If instead $k_1 < k_2$, \[\frac{10^5 - 1}{10^5} < \frac{d_{k_1} + \dots + d_{k_2 - 1}}{d_{k_1} + \dots + d_{k_2}} < 1. \; \; \; \; \; \; \; \; \; \; \blacksquare\] The conclusion follows. \[1 < \frac{d_{k_2} + \dots d_{k_1}}{d_{k_2 + 1} + \dots d_{k_1}} \]this is incorrect it is less than one not bigger.

Solved with Th3Numb3rThr33. Assume for contradiction there is an exception $S$, a set of $2000$ distinct real numbers. Increment all elements of $S$ by a sufficiently large positive number so that all elements of $S$ are nonnegative. Then \[S=\{a_1,a_1+a_2,\ldots,a_1+a_2+\cdots+a_{2000}\}\]for positive real numbers $a_1$, $a_2$, $\ldots$, $a_{2000}$. Let $a_i=\min(a_1,\ldots,a_{2000})$. Let $T$ be the multiset of $|a-b|$ for $a,b\in S$. If we sort the elements of $T$ in increasing order, for any consecutive $p,q\in T$, we have $q/p\ge1+10^{-5}$. There are $\tbinom{2000}2=1999000$ elements in $T$, so \[\frac{a_1+\cdots+a_{2000}}{a_i}=\frac{\max T}{\min T}\ge\left(1+10^{-5}\right)^{1998999}>2^{19}>500000.\] On the other hand, note that \begin{align*} (a_1+\cdots+a_{i-1})\left(1+10^{-5}\right)&\le a_1+\cdots+a_i\\ (a_{i+1}+\cdots+a_{2000})\left(1+10^{-5}\right)&\le a_i+\cdots+a_{2000}. \end{align*}Adding these together, we have \begin{align*} (a_1+\cdots+a_{2000}-a_i)\left(1+10^{-5}\right)\le a_1+\cdots+a_{2000}+a_i\\ \implies\frac{a_1+\cdots+a_{2000}}{a_i}\le\frac{2+10^{-5}}{10^{-5}}=200001, \end{align*}contradiction.

Solution from Twitch Solves ISL: WLOG, we denote the set of $2000$ real numbers by \[ S = \left\{ 0 = x_1 < x_2 < \dots < x_{2000} = 1 \right\}. \]We proceed by contradiction and assume this is not the case. Claim: There exists $0 \le x < y \le 1$ in $S$ such that \[ x-y < \left( \frac{1}{1 + 10^{-5}} \right)^{\binom{2000}{2}-1}. \]Proof. Our contradiction hypothesis assumes that any two differences of elements in $S$ differ by at least a factor of $1 + 10^{-5}$. Since the largest difference is $1-0 = 1$ and there are $\binom{2000}{2}$ differences, the conclusion follows. $\blacksquare$ We let $0 \le x<y \le 1$ denote the elements with smallest difference. Let $\rho = x-y$. An arithmetic calculation shows that $\rho < 3 \cdot 10^{-9}$. [asy][asy] size(8cm); draw( (0,0)--(1,0) ); dot("$0$", (0,0), dir(-90), blue); dot("$1$", (1,0), dir(-90), blue); dot("$x$", (0.63,0), dir(-90), red); dot("$y$", (0.67,0), dir(-90), red); label("super close", (0.65,0), dir(90), red); [/asy][/asy] Note that we have If $x > 0.499$ then \[ 0 < \frac{y-0}{x-0} - 1 < \frac{\rho}{0.499} < 10^{-5} \]as needed. If $ y < 0.501$ then similarly \[ 0 < \frac{1-x}{1-y} - 1 < \frac{\rho}{0.499} < 10^{-5} \]as needed.

Solved with MortemEtInteritum, ApraTrip, and tree_3. I might've forgotten a few others: if I did please PM me and I'll fix it. First note that the condition can be rewritten as $\frac{99999}{100000} < \frac{a-b}{c-d} < \frac{100001}{100000}$. Now, denote the set as $\{x_1,x_2,\ldots,x_{2000}\}$, where $0=x_1<x_2<\cdots x_{2000}$ (we can do this because we can shift the sequence without changing the result). Also, let the $\binom{2000}{2}=1999000$ differences $x_i-x_j$ with $i>j$ be $d_1\leq d_2\leq \cdots\leq d_{1999000}$. Further, note that we can scale the sequence (without changing the result), so WLOG let $d_1=1$. We now proceed with contradiction: we will attempt to make the condition false, and reach a contradictory statement. Suppose FTSOC such that there does not exist $i,j$ with $i \neq j$ such that $\frac{99999}{100000} < \frac{d_i}{d_j} < \frac{100001}{100000}$. This would mean that, for $j>i$, we have $d_j>\frac{100000}{99999}d_i$. In particular, we have $d_{i+1}>\frac{100000}{99999}d_i$. Combining this with $d_1=1$, we find that $d_n>\left(\frac{100000}{99999}\right)^{n-1}$. In particular, $d_{1999000}>\left(1+\frac{1}{99999}\right)^{1998999}$. However, we have $d_{1999000}=x_{2000}-x_1=x_{2000}$, so $x_{2000} > \left(1+\frac{1}{99999}\right)^{1998999}$. We note that $19(99999)<1998999<20(99999)$, so: \begin{align*} x_{2000}&>\left(1+\frac{1}{99999}\right)^{1998999}\\ &>\left(1+\frac{1}{99999}\right)^{19*99999}\\ &>\left(1+\frac{99999}{99999}\right)^{19}\\ &=2^{19}\\ &=524288 \end{align*}We use Bernoulli to get from the second line to the third. Now, since $d_1=1$, there exists an $n$ such that $x_n-x_{n-1}=1 \implies x_n=x_{n-1}+1$ (the reason they have to be consecutive is because due to our ordering restriction, we cannot have non-consecutive numbers which result in a minimal difference). We now consider the following cases: Case 1: $x_n>200000$ Then, we find that $\frac{x_n-x_1}{x_{n-1}-x_1}=\frac{x_n}{x_n-1}$. Since $x_n>200000$, this falls in the range $\left(\frac{99999}{100000}, \frac{100001}{100000}\right)$, which is a contradiction. Case 2: $x_n\leq 200000$. Note that since $x_{2000}>524288$, this implies that $n \neq 2000$. Then, we find that $\frac{x_{2000}-x_n}{x_{2000}-x_{n-1}}=\frac{x_{2000}-x_n}{x_{2000}-x_n+1}$. Since $x_{2000}>524288$ and $x_n \leq 200000$, we have $x_{2000}-x_n>300000$, so this falls in the range $\left(\frac{99999}{100000}, \frac{100001}{100000}\right)$, which is a contradiction as well. Combining these cases, we see that the problem statement must be satisfied, as desired. $\blacksquare$

Solved with nukelauncher. Let $n=2000$ and $m=10^5$. Suppose for the sake of contradiction that there existed a set of 2000 distinct real numbers such that for every two pairs $a>b$, $c>d$ satisfying $a\neq c$ or $b\neq d$, then \[\left\lvert \frac{a-b}{c-d}-1 \right\rvert\geq\frac1m.\]WLOG we may shift the set of 2000 real numbers such that the minimum is 0, then scale so the maximum is 1; both these operations leave all values of $\tfrac{a-b}{c-d}$ unchanged. Now take real numbers $x_0=0$, $x_1,x_2,\dots,x_{n-1}$ such that $x_0+\dots+x_{n-1}=1$ and the set is \[ 0=x_0 \le x_1 \le x_1+x_2 \le \cdots\le x_1+\cdots+x_{n-1}=1.\] Claim: $\min(x_1,x_2,\dots,x_{n-1})\geq\tfrac1{2m}$. Proof: Suppose not, and let $x_i<\tfrac1{2m}$. Since $x_0+\dots+x_{i-1}+x_{i+1}+\dots+x_{n-1}>1-\tfrac1{2m}$ this means that $\max(x_0+\dots+x_{i-1},x_{i+1}+\dots+x_{n-1})>\tfrac12-\tfrac1{4m}$. WLOG suppose $x_0+\dots+x_{i-1}>\tfrac12-\tfrac1{4m}$. Now take $a-b=x_0+\dots+x_{i-1}>\tfrac12-\tfrac1{4m}$ and $c-d=x_0+\dots+x_i$. Let $f(x)=\tfrac x{x+x_i}$. Since $f(x)$ is an increasing function of $x$ on the positive real numbers, we have \begin{align*} \frac{a-b}{c-d}&=f(x_0+\dots+x_{i-1})>f\left( \frac12-\frac1{4m} \right)\\ &=\frac{\frac12-\frac1{4m}}{\frac12-\frac1{4m}+x_i}>\frac{\frac12-\frac1{4m}}{\frac12+\frac1{4m}}=\frac{2m-1}{2m+1}>1-\frac1m. \end{align*}We also have that $\tfrac{a-b}{c-d}<1$, thus the above implies \[\left\lvert \frac{a-b}{c-d}-1 \right\rvert<\frac1m,\]contradiction. $\blacksquare$ Claim: $\min(x_1,x_2,\dots,x_{n-1})\leq\left( 1-\frac1m \right)^{\binom n2-1}$. Proof: Order the $\tbinom{n}{2}$ set of contiguous sums of $(x_1,\ldots,x_{n-1})$ \begin{align*} &x_1,x_2,\dots,x_{n-1},\\ &x_1+x_2,x_2+x_3,\dots,x_{n-2}+x_{n-1},\\ &x_1+x_2+x_3,\dots,x_{n-3}+x_{n-2}+x_{n-1},\\ &x_1+x_2+x_3+x_4,\dots,x_{n-4}+x_{n-3}+x_{n-2}+x_{n-1},\\ &\vdots\\ &x_1+\dots+x_{n-1}. \end{align*}in increasing order. Call this set of numbers $A=\left\{ a_1<a_2<\dots<a_{\tbinom n2} \right\}$. Note that by construction, every $a_i$ can be written as $a-b$ for some $a,b$ in the original set; therefore, by our hypothesis about the original set, we have \[\left\lvert \frac{a_i}{a_{i+1}}-1 \right\rvert\geq\frac1m.\]Since $a_i<a_{i+1}$ this becomes \[\frac{a_i}{a_{i+1}}\leq 1-\frac1m\implies a_i\leq \left( 1-\frac1m \right)a_{i+1}.\]Chaining this inequality by multiplying them for $i=1,\dots,\tbinom n2-1$ and noting that \[a_1=\min(x_1,x_2,\dots,x_{n-1}),\quad a_{\binom n2}=x_1+\dots+x_{n-1}=1\]gives \[\min(x_1,x_2,\dots,x_{n-1})\leq\left( 1-\frac1m \right)^{\binom n2-1},\]as desired. $\blacksquare$ Now, combining the two claims yields \[\left( 1-\frac1m \right)^{\binom n2-1}\geq\frac1{2m}\implies 2m\geq \left( 1+\frac1{m-1} \right)^{\binom n2-1}.\]Finally, we have \begin{align*} 2\cdot10^5&\geq\left( 1+\frac1{m-1} \right)^{\binom n2-1}>\left( \left( 1+\frac1m \right)^m \right)^{\frac{\binom n2-1}m}\\ &>\left( 1+\frac mm \right)^{\frac{1000\cdot 1999-1}{10^5}}>2^{19}>2\cdot 10^5, \end{align*}contradiction. Thus, no set exists which violates the property in the problem, as desired.

Assume the contrary. Let the 2000 reals be $a_1 < a_2 <\cdots < a_{2000}.$ Consider all $\binom{2000}{2} = 1999000$ possible pairwise differences, where $r_1 < r_2 < \cdots < r_{1999000}.$ By assumption, $r_{i+1} \ge \frac{100000r_i}{99999} = kr_i$ for brevity's sake. As a result, we have $$a_{2000} - a_1 = r_{1999000} \ge k^{1998999} r_1.$$Moreover, note that $$a_{2000}-a_1 = \sum_{i=2}^{2000} a_i-a_{i-1} = r_1(k^{x_1} + k^{x_2} + \cdots + k^{x_{1999}}) \stackrel{\text{Jensen's}}{\le} 1999r_1k^{1998999/1999} \le 1999r_1 k^{1000}.$$As a result, it must be true that $k^{1997999} \le 1999.$ However, note from Binomial Theorem, $$\left(1 + \frac{1}{99999}\right)^{1997999} \ge \left(1+\frac{99999}{99999}\right)^{1997999/99999} > 2^{19} > 1999,$$Contradiction. $\blacksquare$

A lot shorter than some other solutions but I think it's near-equivalent to v_Enhance's solution so it should be fine. Assume the contrary. Rotate and scale so that the smallest number is $0$ and the largest number is $1$. Notice that there are $\binom{2000}{2} = 1999000$ possible positive differences that can be achieved. Since the largest difference is $1$, and we can't have two positive differences so that the smaller one is more than $1-\frac{1}{100000}$ times the other, the smallest positive difference must be at most $$\left(1-\frac{1}{100000}\right)^{1999000-1}< \left(\left(1-\frac{1}{100000}\right)^{100000}\right)^{19}< \left(\frac{1}{e}\right)^{19} <\frac{1}{2^{19}} < \frac{1}{500000}.$$Let $c$ and $d$ be the two numbers in the set such that $c-d$ is this smallest positive difference. Then if the average of $c$ and $d$ is greater than or equal to $\frac{1}{2}$, consider $\frac{c}{d}$, otherwise consider $\frac{1-d}{1-c}$. This number must be greater than $1$ but at most $$\frac{\frac{1}{2}+\frac{1}{10^6}}{\frac{1}{2}-\frac{1}{10^6}} < 1 + \frac{1}{100000},$$contradiction.

Let the set be $S$. Wlog, shift, and scale $S$ so that the smallest element is $1$ and the largest is $2$.Note that there are $N:=\binom{2000}{2}$ different possible differences which we denote as $d_1<d_2< \cdots <d_N$. Let $x_n$ denote $\log_2(d_n)$. Thus, $1=x_1,x_2,...,x_{N} =2$ divides the interval $[0,1]$ into $\binom{2000}{2}+1$ disjoint subintervals. Let $[x_k,x_{k+1}]$ be the smallest such interval. Since there are $N+1$ intervals then $x_{k+1}-x_k \leq \frac{1}{N+1}$. So, \[ \frac{d_{k+1}}{d_k} = \frac{2^{x_{k+1}}}{2^{x_k}} = 2^{x_{k+1}-x_k} \leq 2^{\frac{1}{N+1}} \]Thus, all that is left to do is show that $2^{\frac{1}{N+1}}<1+\frac{1}{100000}$. By Bernoulli's inequality \[ \left (1+\frac{1}{100000} \right )^{1999001} > 1+\frac{1999001}{100000}>2\] $\blacksquare$ Remark: If my solution is correct, this bound is actually quite weak. By the same proof, we can see that $1999001$ could work instead of $100000$

Absolutely garbage (sorry). Suppose otherwise. Sort the $N=\binom{2000}{2}$ differences as $d_1<d_2<\cdots<d_N$ and the elements themselves as $a_1<a_2<\dots<a_{2000}$. Note that the condition in the problem implies that $d_{i+1}\ge d_i\left(\frac{10^5}{10^5-1}\right)$. Also note that, by setting $(a,b,c,d)=(a_{2000},a_2,a_{2000},a_1)$, we get $\frac{d_1}{d_N}\ge \frac{1}{10^5}$. Then note that we also have $$d_N\ge d_1\left(\frac{10^5}{10^5-1}\right)^{N-1}$$so to obtain a contradiction it suffices to show that \[\left(1+\frac{1}{10^5-1}\right)^{N-1}>10^5.\]However, note that Bernoulli's Inequality gives \[\left(1+\frac{1}{10^5-1}\right)^{N-1}>2^{\tfrac{N-1}{10^5-1}}>2^{19}>10^5\]so we are done. $\blacksquare$

Ratio of smallest (A) to biggest: $\left(1+\frac{1}{99999}\right)^{1999000} > 2^{19.99} > 10^6 > 2 \cdot 10^5$ Let the smallest partition the biggest into two intervals, then take the larger one (B). The ratio $\frac{B}{B+A}$ is as desired.

Assume on the contrary. WLOG, suppose the numbers are $a_1<a_2<\dots < a_{2000}$. Then, consider the $\tbinom{2000}{2}=1999000$ differences $a_i-a_j$ for some $i>j$. Call these differences, in order, $d_1<d_2<\dots<d_{1999000}$. If $d_i>\tfrac{99999}{100000}d_{i+1}$ then \[ \left| \frac{d_i}{d_{i+1}} - 1 \right|< \frac{1}{100000}\]which is a contradiction. Now, this implies \[d_{1999000}\ge d_1 \left(\frac{100000}{99999}\right)^{1999000}\]Let $e_i$ denote $a_{i+1}-a_i$. Note that the largest difference must be $a_{2000}-a_1=e_1+e_2+\dots+e_{1999}$. Let $d_m$ be the maximum of the $e$s, and $d_1$ is the minimum of the $e$s. We have \[d_1 \left(\frac{100000}{99999}\right)^{1999000} \le d_{1999000} \le 1999d_m\]so $d_m \ge cd_1$ where $c=\tfrac{\left(\frac{100000}{99999}\right)^{1999000}}{1999}$. Let $e_m$ be the minimum $e$ and $e_M$ be the maximum $e$. If $m<M$ then \[\left|\frac{e_m+e_{m+1}+\dots + e_{M}}{e_{m+1}+e_{m+2}+\dots + e_{M}}-1\right| < \frac{e_m}{e_M} \le \frac{1}{c}\]and similarly if $m>M$ then \[\left|\frac{e_{m-1}+e_{m}+\dots + e_{M}}{e_{m}+e_{m+1}+\dots + e_{M}}-1\right| < \frac{e_m}{e_M} \le \frac{1}{c}\]So $c<100000$ which is false.

The problem essentially says, if we have 2000 real numbers, and we compute all of the positive differences between them, and sort the results in increasing order, then some term is less than $1.00001$ times the previous. Assume FTSOC that each multiplier is at least $1.00001$. WLOG in the original sequence of 2000 numbers, the smallest is $0$, and scale so that the smallest difference between two terms is $1$. Suppose the sequence in increasing order is $$0,a_1,a_2\dots ,a_{1999}.$$Now, since $a_{1999}-0$ is the largest difference, we have $$a_{1999}\geq 1.00001^{{2000\choose 2}-1}=1.00001^{1,998,999}>1.00001^{1,900,000}>2^{19}>500,000.$$ Suppose the two terms that attain the smallest difference is $a_i$ and $a_{i}+1$ since we did WLOG the difference is 1 (for niceness $a_0=0$ here). Then, by our FTSOC assumption, $$\frac{a_{1999}-a_i}{a_{1999}-a_i-1}\geq \frac{100001}{100000},$$which requires $$a_{1999}-a_i\leq 100000.$$Thus, $$a_i\geq 400000.$$However, we then clearly have $$\frac{a_i+1-0}{a_i}<1.00001,$$contradiction. remark- this problem can actually be "solved" (meaning getting the general solution method) in your head extremely quickly. The only real idea used here is "assume not, then exponential forces two terms to be really close to each other, and two terms being close causes close differences", the rest is just formality.