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.*;

public class Main {
    static long[] base = new long[61];
    static final int MOD = 1000000007;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        base[0] = 1;
        for (int i = 1; i < base.length; i++) {
            base[i] = base[i - 1] * 2 % MOD;
        }
        boolean[][] arr = new boolean[n + 1][61];
        for (int i = 1; i <= n; i++) {
            arr[i] = getArray(sc.nextLong());
        }
        long ans = 0;
        int[][] sums = new int[n + 1][61];
        for (int i = 0; i < 61; i++) {
            for (int j = 1; j <= n; j++) {
                sums[j][i] = sums[j - 1][i];
                if (arr[j][i]) {
                    sums[j][i]++;
                } 
                if (arr[j][i]) {
                    ans += (j - sums[j][i]) * base[i];
                    ans %= MOD;
                } else {
                    ans += sums[j][i] * base[i];
                    ans %= MOD;
                }
                
            }
        }
        System.out.println(ans);
    }
    
    static boolean[] getArray(long x) {
        boolean[] ans = new boolean[61];
        for (int i = 0; i <= 60; i++) {
            ans[i] = x % 2 == 1;
            x /= 2;
        }
        return ans;
    }
}