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<string>
#include<vector>
#include<algorithm>
#include<cmath>

using namespace std;

#define rep(i, n) for(int i = 0; i < n; i++)




// a^n mod を計算する
long int modpow(long int a, long int n, long int mod) {
    long int res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = (a * a) % mod;
        n >>= 1;
    }
    return res;
}

long int modcomb(long int a, long int b, long int mod){
    long int res = 1;
    for(long int i = a; i > a - b; i--){
        res *= i % mod;
        res %= mod;
    }
    res %= mod;
    for(long int i = 1; i <= b; i++){
        long int i_inv = modpow(i, mod - 2, mod);
        res *= i_inv % mod;
        res %= mod;
    }
    res %= mod;
    // cout << res << endl;
    return res;
}



int main(){
    long int mod = 1000000007;
    long int n;
    cin >> n;
    long int a[n];
    rep(i, n) cin >> a[i];
    // 60桁分の1の数
    long int num[61];
    fill(num, num + 61, 0);
    rep(i, n) rep(j, 61){
        if (((a[i] >> j) & 1) == 1){
            num[j]++;
        }
    }
    // rep(i, 61) cout << num[i] << " ";
    // cout << endl;
    // 2のi乗の数がa[i]
    long int ans[61];
    fill(ans, ans + 61, 0);
    rep(i, 61){
        long int num1 = num[i];
        long int num0 = n - num1;
        ans[i] = num0 * num1;
    }
    // rep(i, 60) cout << ans[i] << " ";
    // cout << endl;   
    long int sm = 0;
    long int bin = 1;
    // int count = -1;
    rep(i, 61){
        // count++;
        sm += bin * ans[i];
        sm %= mod;
        bin *= 2;
        bin %= mod;
    }
    // cout << count << endl;
    cout << sm << endl;
}