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>
using namespace std;

const long long mod = 1000000007;
 
int main() {
  long long N; cin >> N;
  vector<long long> cnt(61);
  
  vector<long long> f(61);
  f.at(0) = 1;
  for(int i = 1; i < 61; ++i){
    f.at(i) = f.at(i-1) * 2 % mod;
  }
  
  long long ans = 0;

  for(int i = 0; i < N; ++i){
    long long A; cin >> A;
    bitset<64> x(A);
    for(int j = 0; j < 61; ++j){
      if(x.test(j)) cnt.at(j)++;
    }
  }
  
  for(int i = 0; i < 61; ++i){
    ans += cnt.at(i) * (N - cnt.at(i)) % mod * f.at(i) % mod;
    ans %= mod;
  }
  
  cout << ans << endl;
}