A boy have 32 cards out of which 10 cards, each of blue, green and red colours, have denominations as \( \left\{2^{1}, 2^{2}, \ldots, 2^{10}\right\} \) and one black and one white each has value unity. The number of ways in which the boy can get a sum of 2012 if he can choose any number of cards, is
(A) \( (1001)^{2} \)
(B) \( (1002)^{2} \)
(C) \( (1007)^{2} \)
(D) \( \quad(1111)^{2} \)
