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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
int main(){
  int N, z=1e9+7;
  cin >> N;
  vector<long long> memo(N);
  int bit=0, temp=0;
  for(int i=0; i<N; i++){
    long long a;
    cin >> a;
    memo.at(i)=a;
    if(a!=0){
    temp=log2(a)+1;
    bit=max(bit,temp);
    }
  }
  vector<long long> zero(bit,0), one(bit,0);
  for(int i=0; i<N; i++){
    long long a=memo.at(i);
    for(int j=0; j<bit; j++){
      if(a>>j &1){
        one.at(j)++;
//      cout<<"i:"<<i<<" j:"<<j<<" a>>j1:"<<(a>>j)<<endl;
      }
      else{
        zero.at(j)++;
//      cout<<"i:"<<i<<" j:"<<j<<" a>>j0:"<<(a>>j)<<endl;
      }
    }
  }
  long long ans=0;
  for(int i=bit-1; i>=0; i--){
    ans*=2;
    ans+=zero.at(i)*one.at(i);
  }
  ans%=z;
  cout << ans<< endl;
}