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: Runtime Error)

// ABC 147-D
// 2019.12.8 bal4u

#include <stdio.h>
#include <ctype.h>

typedef long long ll;

#if 1
#define gc() getchar_unlocked()
#else
#define gc() getchar()
#endif

ll in() {   // 非負整数の入力
	ll n = 0; int c;
	do c = gc();
	while (isspace(c));
	do n = 10 * n + (c & 0xf); while ((c = gc()) >= '0');
	return n;
}

#define M 1000000007
int N;
ll A[100005];

int main()
{
	int i, j, ans;
	
	N = (int)in();
	for (i = 0; i < N; ++i) {
		A[i] = in();
	}
	ans = 0;
	for (i = 0; i < N; ++i) {
		for (j = i+1; j < N; ++j) {
			ans = ((ll)ans + (A[i]^A[j])) % M;
		}
	}
	printf("%d\n", ans);
	return 0;
}