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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

const long long MOD = 1000000007;
const int INF = 1e9;

bool BitFlag(long long n, int i){
  if(n&(1<<i)) return 1;
  else return 0;
}

int main(){
  long long n;
  const long long max_n = 3 * 1e5;
  long long a[max_n + 1];
  cin >> n;
  for(long long i = 0; i < n; i++) cin >> a[i];

  long long num[62] = {};
  for(long long i = 0; i <= 61; i++){
    for(long long j = 0; j < n; j++){
      if(a[j]&(1LL<<i)) num[i] += 1;
    }
  }
  
  long long ans = 0;
  for(long long i = 0; i <= 61; i++){
    long long now = (1LL << i);
    ans += num[i] * (n-num[i]) * now;
    ans %= MOD;
  }
  cout << ans << endl;

  return 0;
}