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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>

int main(int argc, char **argv)
{
	int N;
	scanf("%d", &N);
	uint64_t *a = malloc(sizeof(uint64_t) * N);
	for (int i = 0; i < N; i++)
		scanf("%ld", &a[i]);
	uint64_t sum = 0;
	int sz[60] = {0};
	int so[60] = {0};
	for (int b = 0; b < 60; b++) {
		uint64_t c = (uint64_t)1 << b;
		for (int i = 0; i < N; i++) {
			if (a[i] & c)
				++so[b];
			else
				++sz[b];
		}
		fprintf(stderr, "%d: %d %d\n", b, sz[b], so[b]);
		if (sz[b] && so[b]) {
			sum += c * sz[b] * so[b];
			sum %= 1000000007;
		}
	}
	printf("%ld\n", sum % 1000000007);
	return 0;
}