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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>

int xor (int, int);
int main()
{
    int a[300000];
    int n,i,j,sum=0;

    scanf("%d", &n);
    for (i = 0; i < n; i++)
    {
        scanf("%d" ,&a[i]);
    }
    for (i = 0; i < n-1;i++){
        for (j = i + 1; j < n;j++){
            sum += xor(a[i], a[j]);
        }
    }
        printf("%d\n", sum/(1000000007));

    return 0;
}

int xor (int a, int b) {
    int x = 0, y = 0, i, j, t = 2, c, d,sum = 0;

    while(1)
    {
        c = a % t;
        d = b % t;
        if(c!=d)
        {
            sum += t / 2;
        }
        a -= c;
        b -= d;
        t = t * 2;
        if(0==a||0==b)
        {
            sum = sum + a + b;
            break;
        }
    }
    return sum;
}