Problem:
Score : 400 points

Problem Statement

We have N integers. The i-th integer is A_i.

Find \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

What is \mbox{ XOR }?

The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:

When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.

For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)

Constraints

2 \leq N \leq 3 \times 10^5

0 \leq A_i < 2^{60}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the value \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

Sample Input 1

3
1 2 3

Sample Output 1

6

We have (1\mbox{ XOR } 2)+(1\mbox{ XOR } 3)+(2\mbox{ XOR } 3)=3+2+1=6.

Sample Input 2

10
3 1 4 1 5 9 2 6 5 3

Sample Output 2

237

Sample Input 3

10
3 14 159 2653 58979 323846 2643383 27950288 419716939 9375105820

Sample Output 3

103715602

Print the sum modulo (10^9+7).

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

class Main {
    public static void main(String[] args) {
        final long MOD = 1_000_000_007L;
        final Scanner sc = new Scanner(System.in);
        final int N = Integer.parseInt(sc.next());
        int B[][] = new int[N][60];
        for (int i=0; i<N; i++) {
            long A =  Long.parseLong(sc.next());
            for (int j=0; j<60; j++) {
                B[i][j] = (int)(A % 2);
                A /= 2;
                if (A==0)
                    break;
            }
        }
        sc.close();

        long sum = 0L;
        for (int j=0; j<60; j++) {
            long cnt = 0;
            for (int i=0; i<N; i++) {
                cnt += B[i][j];
            }
            sum += cnt * (N - cnt) * Math.pow((double)2, (double)j) % MOD;
            sum %= MOD;
        }
        System.out.println(sum);
    }
}