Different Bits Sum Pairwise

We define f(X, Y) as number of different corresponding bits in binary representation of X and Y. For example, f(2, 7) = 2, since binary representation of 2 and 7 are 010 and 111, respectively. The first and the third bit differ, so f(2, 7) = 2.

You are given an array of N positive integers, A1, A2 ,…, AN. Find sum of f(Ai, Aj) for all pairs (i, j) such that 1 ≤ i, j ≤ N. Return the answer modulo 109+7.

For example,

A=[1, 3, 5]

We return

f(1, 1) + f(1, 3) + f(1, 5) + 
f(3, 1) + f(3, 3) + f(3, 5) +
f(5, 1) + f(5, 3) + f(5, 5) =

0 + 1 + 1 +
1 + 0 + 2 +
1 + 2 + 0 = 8

public class Solution {
    public int cntBits(int[] A) {
        int n = A.length;
        long dist = 0;
        for (int i = 0; i < 31; i++) {
            int oneCount = 0;
            for (int j = 0; j < n; j++) 
                oneCount += (A[j] >> i & 1);
            int zeroCount = n - oneCount;
            dist = (dist + 2L * oneCount * zeroCount) % 1000000007;
        }
        return (int) dist;
    }
}