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 <bits/stdc++.h>
using namespace std;
long long mod=1e9+7;
int main() {
    int n;
    cin>>n;
    long long a[n];
    int b[n][62];
    int c[62];
    long long ans[62];
    memset(c,0,sizeof(c));
    memset(ans,0,sizeof(ans));
    memset(b,0,sizeof(b));
    for(int i=0;i<n;i++){
        cin>>a[i];
        long long k=a[i];
        int j=0;
        while(k){
            b[i][j]=k%2;
            c[j]+=b[i][j];
            k/=2;
            j++;
        }
    }
    for(int i=0;i<n;i++){
        for(int j=0;j<62;j++){
            c[j]-=b[i][j];
            if(b[i][j]==1){
                ans[j]+=n-i-1-c[j];
            }
            else{
                ans[j]+=c[j];
            }
            ans[j]%=mod;
        }
    }
    long long z=0,yo;
    for(int i=0;i<62;i++){
        yo=1;
        for(int j=0;j<i;j++){
        yo*=2;
            yo%=mod;
        }
        yo*=ans[i];
        yo%=mod;
        z+=yo;
        z%=mod;
    }
    cout<<z%mod;
    return 0;
}