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 <iostream>
#include <iomanip>
#include <vector>
#include <cmath>
#include <algorithm>
#include <bitset>
using namespace std;

const int MOD = 1e9 + 7;

long long mod_pow(int a, int n) {
    long long r = 1;
    for (int i = 0; i < n; i++) {
        r = (r * a) % MOD;
    }
    return r;
}

int main() {
    int N; cin >> N;
    vector<long long> As(N);
    vector<pair<int, int> > Keta(62, {0, 0});
    for (int i = 0; i < N; i++) {
        cin >> As[i];
        std::stringstream ss;
        ss << std::bitset<62>(As[i]);
        std::string s = ss.str();
        reverse(s.begin(), s.end());
        for (int j = 0; j < 62; j++) {
            if (s[j] - '0' == 0) {
                Keta[j].first++;
            } else {
                Keta[j].second++;
            }
        }
    }
    
    long long ans = 0;
    for (int i = 0; i < 60; i++) {
        ans = (ans + mod_pow(2, i) * ((Keta[i].first * Keta[i].second) % MOD)) % MOD;
    }
    cout << ans << endl;
}