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 _GLIBCXX_DEBUG
#define rep(i, n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;

int modpow(int a, int b, int mod){
    int x = 1;
    for (; b > 0; b >>= 1){
        if(b & 1) x = (ll)x * a % mod;
        a = (ll)a * a % mod;
        x %= mod;
    }
    return x;
}

int main(){
    int N;
    cin >> N;
    vector<int> cnt(60);
    vector<ll> A(N);
    ll ans = 0;
    rep(i, N) cin >> A[i];
    rep(i, N) rep(j,60) if(A[i]>>j & 1) cnt[j]++;
    int mod = 1e9+7;
    rep(i,60){
        if(cnt[i] == 0) continue;
        ll x = ((ll)modpow(2,i,mod) * (N-cnt[i])) % mod;
        x = x * cnt[i] % mod;
        ans += x;
        ans %= mod;
    }
    cout << ans << endl;
    return 0;
}