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;
#define rep(i, n) for(int i = 0; i < n; ++i)
#define repab(i, a, b) for(int i = a; i < b; ++i)
#define rrep(i, n) for(int i = n-1; i >= 0; --i)
typedef unsigned long long ll;

ll mod = 1000000007;

int main(){
  	ll n; cin>>n;
  	ll count = 0;
  	ll *arr = new ll[n];
  	ll *dig = new ll[n];
  	ll mu = 1;
  	ll cum = 0;
  	ll oz[2] = {};
  	rep(i, n){
    	cin>>arr[i];
      	dig[i] = 0;
    }
  	rep(idx, 64){
      	cum = 0;
      	oz[0] = 0;
      	oz[1] = 0;
    	rep(i, n){
          dig[i] = arr[i] % 2;
          arr[i] >>= 1;
        }
      	oz[dig[0]] = 1;
      	repab(i, 1, n){
        	cum += oz[dig[i]?0:1];
          	oz[dig[i]] += 1;
        }
        count += (cum * (mu % mod))% mod ;
      	mu *= 2;
      	count %= mod;        
    }
  	cout<<count;
	return 0;
}