#include<bits/stdc++.h>
using namespace std;

const int maxn=3e5+6,p=1e9+7;
long long a[maxn],ans;
int t,cnt,n;

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i) {
		scanf("%lld",&a[i]);
	}
	for(int j=0;j<=60;++j) {
		cnt=0;
		for(int i=1;i<=n;++i)
			if(a[i]&(1ll<<j)) cnt++;
		ans+=((((1ll<<j)%p*cnt)%p)*(n-cnt))%p;
		ans%=p;
	}
	printf("%lld",ans%p);
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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).

Examples

Input

3
1 2 3


Output

6


Input

10
3 1 4 1 5 9 2 6 5 3


Output

237


Input

10
3 14 159 2653 58979 323846 2643383 27950288 419716939 9375105820


Output

103715602

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

const int maxn=3e5+6,p=1e9+7;
long long a[maxn],ans;
int t,cnt,n;

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i) {
		scanf("%lld",&a[i]);
	}
	for(int j=0;j<=60;++j) {
		cnt=0;
		for(int i=1;i<=n;++i)
			if(a[i]&(1ll<<j)) cnt++;
		ans+=((((1ll<<j)%p*cnt)%p)*(n-cnt))%p;
		ans%=p;
	}
	printf("%lld",ans%p);
}
```