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)

/* Coached by rainboy */
#include <stdio.h>

#define N 300000LL
#define L 60LL
#define MD 1000000007LL

void reverse(long long n, char *s) {
	long long i, j;
	char tmp;

	for (i = 0, j = n - 1; i < j; i++, j--)
		tmp = s[i], s[i] = s[j], s[j] = tmp;
}

void insert(long long n, char *s) {
	long long i, j;

	for (i = n - 1, j = L - 1; i >= 0; i--, j--)
		s[j] = s[i];
	for (i = 0; i < L - n; i++)
		s[i] = '0';
}

int main() {
	static char bb[N][L + 1];
	long long n, i, j, k, ans;

	scanf("%lld", &n);
	for (i = 0; i < n; i++) {
		long long a;

		scanf("%lld", &a);
		k = 0;
		while (a > 0) {
			bb[i][k++] = a % 2 + '0';
			a /= 2;
		}
		reverse(k, bb[i]);
		insert(k, bb[i]);
	}
	ans = 0;
	for (j = 0; j < 60; j++) {
		k = 0;
		for (i = 0; i < n; i++)
			if (bb[i][j] == '0')
				k++;
		ans = (ans + k * (n - k) * (1LL << 59 - j)) % MD;
	}
	printf("%lld\n", ans);
	return 0;
}