Let $\Pi_s=a$, from Willson $\bar{{\Pi}_s}=-1/a$. So we have to find min minimal $a-(-1/a)=a+1/a (mod p)$. So let's have a look at the first three cases: 1)$a+1/a=0 (mod p)$ means $a^2=-1 (mod p)$ but for p=3 (mod 4) is -1 not quadratic residue. 2)$a+1/a=1 (mod p)$ means $4a^2-4a=-4 (mod p)$ $(2a-1)^2=-3 (mod p)$, bud for p=11 (mod 12) is -3 not quadratic residue. 3)$a+1/a=2(mod p)$. We can obtain this as *$[(p-1).(p-2).....((p+3)/2)).((p+1)/2)]-[1.2.....((p-3)/2).((p-1)/2)]$ or #$[(p-1).(p-2).....((p+3)/2)).((p-1)/2)]-[1.2.....((p-3)/2).((p+1)/2)]$. From Willson the product of these two brackets in * is -1 and their values mod p differ only in sign, so one is 1 and the other is -1, as well as the brackets in #. But the values of first bracket in * and # differ in sign, so one of * and # works.