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: Wrong Answer)

import java.util.Arrays;
import java.util.Scanner;

public class Main {
    static final long MOD = 1000000007;

    public Main() {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();

        long[] A = new long[N];

        for (int i = 0; i < N; i++) {
            A[i] = scanner.nextLong();
        }

        long R = 0;

        for (int D = 0; D < 60; D++) {
            int zNum = 0, oNum = 0;
            long B = 1L << D;
            for (int i = 0; i < N; i++) {
                if ( ( A[i] & B ) == B ){
                    oNum++;
                } else {
                    zNum++;
                }
            }

            R = (R + B * (oNum * zNum % MOD) % MOD) % MOD;

        }

        System.out.println(R);
    }


    long modPow(long n){
        long R = 1L;

        long A = 2;
        while(n > 0){
            if( (n & 1) == 1)
                R = (R * A) % MOD;
            A = (A * A) % MOD;
            n = n >> 1;
        }

        return R;
    }

    public static void main(String[] args) {
        new Main();
    }
}