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: Runtime Error)

/**
 * Title: D - Xor Sum 4
 * Url: https://atcoder.jp/contests/abc147/tasks/abc147_d
 */
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<numeric>
#include<functional>
#include<sstream>
#include<iomanip>

using namespace std;

#define ll long long
#define MOD 1000000007

int main() {
    int n; cin >> n;
    vector<ll> a(n);
    for(int i=0; i<n; i++) cin >> a[i];
    auto max_elm = max_element(a.begin(), a.end());
    int keta = log2(*max_elm);
    vector<vector<int>> m(keta+1, vector<int>(2, 0));
    for(int i=0; i<n; i++) {
        for(ll j=0; j<=keta; j++) {
            if(a[i] & (1ll<<j)) {
                m[j][1] += 1;
            } else {
                m[j][0] += 1;
            }
        }
    }
    ll ans = 0;
    for(int i=0; i<=keta; i++) {
        ans += pow(2, i)*m[i][0]*m[i][1];
        ans %= MOD;
    }
    cout << ans << endl;
    return 0;
}