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;
typedef long long ll;
typedef long double ld;
const ll MAX = 2000000;
const ll MOD = 1000000007;
ll n;

int main() {
    cin >> n;
    ll ketasum[60];
    for(ll jp=0; jp<60; jp++){ketasum[jp] = 0;}
    ll sum = 0;
    for(ll i=0; i<n; i++){
        ll tempA;
        cin >> tempA;
        ll rest = tempA;
        for(ll j=0; rest>0; j++){
            ketasum[j] += rest%2;
            rest = floor(rest/2);
        }
    }
    ll pow = 1;
    for(ll k=0; k<60; k++){
        //cout << ketasum[k] << endl;
        ll temp = ketasum[k] * (n - ketasum[k]);
        sum = (sum + (temp * pow)%MOD)%MOD;
        pow = (pow * 2) % MOD;
    }
    cout << sum << endl;
    return 0;
}