A checkered polygon $A$ is drawn on the checkered plane. We call a cell of $A$ internal if all $8$ of its adjacent cells belong to $A$. All other (non-internal) cells of $A$ we call boundary. It is known that $1)$ each boundary cell has exactly two common sides with no boundary cells; and 2) the union of all boundary cells can be divided into isosceles trapezoid of area $2$ with vertices at the grid nodes (and acute angles of the trapezoids are equal $45^\circ$). Prove that the area of the polygon $A$ is congruent to $1$ modulo $4$.