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 <inttypes.h>

#define MOD_BY UINT32_C(1000000007)

uint32_t add(uint32_t a, uint32_t b) {
	return a + b - MOD_BY * (a + b >= MOD_BY);
}

uint32_t sub(uint32_t a, uint32_t b) {
	return b == 0 ? a : add(a, MOD_BY - b);
}

uint32_t mul(uint32_t a, uint32_t b) {
	uint32_t r = 0;
	while (b > 0) {
		if (b & 1) r = add(r, a);
		a = add(a, a);
		b >>= 1;
	}
	return r;
}

uint32_t N;
uint64_t A[333333];

uint32_t sum[333333];
uint32_t count[333333][64];

int main(void) {
	uint32_t i, j;
	uint32_t answer = 0;
	if (scanf("%" SCNu32, &N) != 1) return 1;
	for (i = 1; i <= N; i++) {
		if (scanf("%" SCNu64, &A[i]) != 1) return 1;
	}
	for (i = 1; i <= N; i++) {
		sum[i] = add(sum[i - 1], (uint32_t)(A[i] % MOD_BY));
		for (j = 0; j < 64; j++) {
			count[i][j] = count[i - 1][j] + ((A[i] >> j) & 1);
		}
	}
	for (i = 1; i < N; i++) {
		uint32_t delta = 0;
		for (j = 0; j < 64; j++) {
			if ((A[i] >> j) & 1) {
				uint32_t one = count[N][j] - count[i][j];
				uint32_t zero = (N - i) - one;
				uint32_t thisDelta = (uint32_t)((UINT64_C(1) << j) % MOD_BY);
				delta = add(delta, mul(thisDelta, sub(zero, one)));
			}
		}
		answer = add(answer, add(sub(sum[N], sum[i]), delta));
	}
	printf("%" PRIu32 "\n", answer);
	return 0;
}