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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
using namespace std;

int main()
{
    int n;
    cin >> n;

    long long a[n];

    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
    }

    int bit[n][60];
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < 60; j++)
        {
            bit[i][j] = a[i] % 2;
            a[i] = a[i] >> 1;
        }
    }


    int count[60] = {};
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < 60; j ++)
        {
            if (bit[i][j] == 1)
            {
                count[j]++;
            }
        }
    }
    // calc
    const long long DIV = 1000000007;
    long long ans = 0;
    long long base = 1;
    for (int i = 0; i < 60; i++)
    {
        //cout << i << ": " << count[i] << ", " << n- count[i] << endl;
        long long temp = (count[i] * base) % DIV;
        ans += (temp * (n - count[i])) % DIV;
        base = (base * 2) % DIV;
    }
    ans = ans % DIV;
    cout << ans << endl;

    return 0;
}