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 <iostream>
#include <vector>
using namespace std;
using ll = long long;
const ll mod = 1000000007;
int main(){
  int n;
  cin >> n;
  vector<ll> a(n);
  for(int i=0;i<n;i++) cin >> a[i];

  /*
  vector<vector<int>> table(n);
  for(int i=0;i<n;i++){
    //各bitを掃き出す
    for(int j=0;j<60;j++){
      ll bit = a[i]>>j & 1;
      table[i].push_back(bit);
    }
  }
  */

  /*
  for(int i=0;i<n;i++){
    for(int j=0;j<60;j++){
      cout<<table[i][j];
    }
    cout << endl;
  }
  */

  ll sum = 0;
  ll bin = 1;
  for(int j=0;j<62;j++){
    int cnt_one = 0;
    for(int i=0;i<n;i++){
      //if(table[i][j] == 1) cnt_one++;
      if(a[i]>>j&1) cnt_one++;
    }
    sum += bin * cnt_one %mod * (n-cnt_one);
    sum %= mod;
    bin *= 2;
    bin %= mod;
  }

  cout << sum << endl;
}