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>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <map>
#include <queue>
#include <bitset>

using ll = long long;
using ull = unsigned long long;
using graph = std::vector<std::vector<ll>>;

using namespace std;

int main() {
    ll N;
    cin >> N;
    unsigned long long A[N];
    for(int i = 0; i < N; i++){
        cin >> A[i];
    }
    ll count[60];
    for(int i = 0; i < 60; i++){
        count[i] = 0;
    }
    for(int i = 0; i < N; i++){
        for(int j = 0; j < 60; j++){
            //cout << A[i] << " ";
            if(A[i] & (1ull << j)){
                //cout << j << endl;
                count[j]++;
            }
        }
    }
    ll MOD = 1000000000 + 7;
    ll result = 0;
    for(int i = 0; i < 60; i++){
        //cout << count[i] * (N - count[i]) * (1 << i) << endl;
        ull tmp = (count[i] * (N - count[i])) % MOD;
        ull tmp2 = (1ull << i) % MOD;
        ull tmp3 = (tmp * tmp2) % MOD;
        result = (result + tmp3) % MOD;
    }
    cout  << result << endl;
}