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 <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep1(i, n) for (int i = 1; i < (n); ++i)
using namespace std;
typedef long long ll;
const ll INF = 1LL << 60;

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    ll n;
    cin >> n;
    vector<vector<ll>> a(n, vector<ll>(60, 0));
    rep(i, n){
        ll num;
        cin >> num;
        ll d = 0;
        while(num){
            if(num%2) a[i][d] = 1;
            num /= 2;
            ++d;
        }
    }

    // rep(i, n) {
    //     rep(j, 60) cout << a[i][j];
    //     cout << endl;
    // }

    vector<ll> bit_num(60, 0);
    rep(i, n){
        rep(j, 60) bit_num[j] += a[i][j];
    }

    // cout << endl;
    // rep(i, 60) cout << bit_num[i];
    // cout << endl;

    ll MOD = 1000000007;
    ll res = 0;
    ll base = 1;
    rep(i, 60){
        res += base%MOD * bit_num[i]%MOD * (n - bit_num[i])%MOD;
        res %= MOD;

        base *= 2;
    }
    cout << res << endl;

    return 0;
}