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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <string>
#include <math.h>
#include <bitset>
#include <vector>
#include <queue>
#include <map>

#define i64 int64_t
#define ff(ii,nn,mm) for(int ii=nn;ii<mm;ii++)
#define sort(vvv) sort(vvv.begin(),vvv.end())
#define rvs(vvv) reverse(vvv.begin(),vvv.end())

using namespace std;

int main() {
	int n;
	int inf = 1000000007;
	cin >> n;
	vector<int> data(n);
	ff(i, 0, n) {
		cin >> data.at(i);
	}
	i64 ans = 0;
	ff(i, 0, 60) {
		i64 z = 1 << i;
		i64 x, y;
		x = 0;//num of 1
		y = n;//num of 0
		ff(j, 0, n) {
			bitset<60> k(data.at(j));
			if (k.test(i)) {
				x++;
				y--;
			}
		}
		i64 a = z * x;
		a %= inf;
		a *= y;
		a %= inf;
		ans += a;
		ans %= inf;
	}
	cout << ans << endl;
	return 0;
}