Hello, struggling sooo much with this question

I hav made an attempt but honestly dont know how to prove the actually 2011^2

Consider the set G of 2011^2 points (p, q) in the plane where p and q are both integers

between 1 and 2011 inclusive. Let

A be any subset of G containing at least 4×2011×square root of 2011

points. Show that there are at least 2011^2

parallelograms whose vertices lie in A and all

of whose diagonals meet at a single point.

Help would b great.

Cheers

I apologise for forgetting to put the 'to the power of' symbols in.