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<stdlib.h>

void tobi(long long num, long long *a);
int power(int m, int n);

int main(void)
{
    long long n,m,a[300000],i,j,counter[60]={0},value[60],nowbi[60],row,ans=0;

    scanf("%lld",&n);
    for ( i = 0; i < n; i++)
    {
        scanf("%lld",a+i);
    }

    for ( i = 0; i < n; i++)
    {
        tobi(a[i],nowbi);
        for ( j = 0; j < 60; j++)
        {
            counter[j]+=nowbi[j];
        }
    }

    for ( i = 0; i < n-1; i++)
    {
        m=n-i-1;
        tobi(a[i],nowbi);
        for ( j = 0; j < 60; j++)
        {
            counter[j]-=nowbi[j];
        }
        for ( j = 0; j < 60; j++)
        {
            if(nowbi[j]==1)
            {
                value[j]=m-counter[j];
            }
            else
            {
                value[j]=counter[j];
            }
        }
        
        row=0;
        for ( j = 0; j < 60; j++)
        {
            row+=value[j]*(power(2,j)%(power(10,9)+7));
        }
        
        ans+=row%(power(10,9)+7);
    }

    printf("%d",ans%(power(10,9)+7));

    return 0;
}

void tobi(long long num, long long *a)
{
    for (int i = 0; i < 60; i++)
    {
        a[i]=(num>>i)&1;
    }
}

int power(int m, int n)
{
    int p=1;
    for (int i = 0; i < n; i++) p*=m;
    return p;
}