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 JavaScript (node.js v5.12) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
#define ALL(c) c.begin(),c.end()
using namespace std;
using ll = long long;
using pii = pair<int,int>;
static const int MOD = 1000000007;

int main(){
    int N; cin >> N;
    vector<ll> A(60);
    for(int i=0; i<N; i++){
        cin >> A[i];
    }
    vector<int> B(60,0);
    for(int i=0; i<N; i++){
        ll a = A[i], bit = 0;
        while(a > 0){
            B[bit] += a%2;
            a = a/2;
            bit++;
        }
    }

    ll ans = 0, n2 = 1;
    for(int i=0; i<60; i++){
        ll addx = (((n2*B[i])%MOD)*(N-B[i]))%MOD;
        ans += addx; ans %= MOD;
        n2 *= 2; n2%MOD;
    }
    cout << ans << endl;
    return 0;
}