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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

#define Q int t; scanf("%d", &t); for(int q=1; q<=t; q++)
typedef unsigned long long int lli;
typedef pair<int, int> pi;


int main()
{
    int n;
    cin >> n;
    lli ones[70] = {0};
    lli zeros[70] = {0};
    while(n--){
        lli x;
        cin >> x;
        for(int i=0; i<63; i++){
            if(x%2){
                ones[i]++;
            }
            else{
                zeros[i]++;
            }
            x /= 2;
        }
    }
    lli sum = 0, mod = 1000000007, mul = 1;
    for(int i=0; i<63; i++){
        lli x = (zeros[i]*ones[i]*mul) % mod;
        sum = (sum+x) % mod;
        mul = (mul*2) % mod;
    }
    //cout << mul << endl;

    cout << sum << endl;

    return 0;
}