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: Accepted)

#include <stdio.h>
#include <math.h>

long modpow(long a, long n, long mod){
	if (n == 0) {
		return 1;
	}
	long answer;
	if (n%2 == 0) {
		answer = modpow(a, n/2, mod);
		answer *= answer;
	} else {
		answer = modpow(a, n-1, mod);
		answer *= a;
	}
	answer %= mod;
	return answer;
}

int main(void) {

  long n;
  scanf("%ld", &n);
  long a[n];
  for (long i = 0; i < n; i++) {
    scanf("%ld", &a[i]);
  }
  long mod = pow(10,9)+7;
  long bit[n][60];
  for (long i = 0; i < n; i++) {
    for (long j = 0; j < 60; j++) {
      bit[i][j] = 0;
    }
  }
  long count_bit[60];
  for (long i = 0; i < 60; i++) {
    count_bit[i] = 0;
  }
  for (long i = 0; i < n; i++) {
    bit[i][0] = a[i];
    for (long j = 1; j < 60; j++) {
      bit[i][j] += bit[i][j-1]/2;
      bit[i][j-1] %= 2;
    }
  }
  for (long i = 0; i < n; i++) {
    for (long j = 0; j < 60; j++) {
      if (bit[i][j] == 1) {
        count_bit[j]++;
      }
    }
  }
  long answer_bit[60];
  long plus;
  long sum = 0;
  for (long i = 0; i < 60; i++) {
    answer_bit[i] = count_bit[i]*(n-count_bit[i]);
    answer_bit[i] %= mod;
    plus = modpow(2,i,mod)*answer_bit[i];
    plus %= mod;
    sum += plus;
    sum %= mod;
  }
  printf("%ld\n", sum);

  return 0;
}