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 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>

long long int xor(long long int a, long long int b) {
  long int i,pow = 0;
  long long int res = 0;
  long long int x;
  while(1) {
    if(a%2 != b%2) {
      x = 1;
      for(i=0; i<pow; i++) x *= 2;
      res += x;
    }
    a /= 2;
    b /= 2;
    pow += 1;
    if(a == 0 && b == 0) break;
  }
  
  return res;
}
 
int main() {
 
  long int i,j,n;
  scanf("%ld", &n);
  long long int *as = malloc(sizeof(long long int) * n);
  for(i = 0; i < n; i++) scanf("%lld", &as[i]);
  
  long long int sum = 0;
  const long long int Y = 1000000007;

  for(i = 0; i < n; i++) 
    for(j = i+1; j < n; j++) 
      sum += xor(as[i], as[j]) % Y;  
  
  printf("%lld", sum % Y);
    
  return 0;
}