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

#define ll long long
#define endl "\n"
#define fastIO ios_base::sync_with_stdio(false)

const ll mod=1e9+7;

int main(){
	fastIO;

	ll t;
	cin>>t;
	while(t--){
		ll n;
		cin>>n;
		ll a[n+1];
		for(ll i=1;i<=n;i++){
			cin>>a[i];
		}
		ll bit[60];
		memset(bit,0,sizeof bit);
		for(ll j=59;j>=0;j--){
			for(ll i=1;i<=n;i++){
				if(a[i]&(1LL<<j)){
					bit[j]++;
				}
			}
		}
		ll ans=0;
		for(ll i=1;i<=n;i++){
			/*ll nd[62];
			memset(nd,0,sizeof nd);
			ll ro[62];
			memset(ro,0,sizeof ro);*/
			ll sum=0;
			for(ll j=59;j>=0;j--){
				ll nd=0;
				if(a[i]&(1LL<<j)){
					nd=(bit[j])%mod;
					ll gun=(1LL<<j)%mod;
					nd=(nd*gun)%mod;
				}
				sum=(sum+nd)%mod;
			}
			for(ll j=59;j>=0;j--){
				ll ro=0;
				if(a[i]&(1LL<<j)){
					ro=n%mod;
					ll gun=(1LL<<j)%mod;
					ro=(ro*gun)%mod;
				}else{
					ro=bit[j]%mod;
					ll gun=(1LL<<j)%mod;
					ro=(ro*gun)%mod;
				}
				ans=(ans+((sum*ro)%mod))%mod;
			}
		}
		cout<<ans<<endl;
	}
}
### Prompt

Please create a solution in cpp to the following problem:

Only a few know that Pan and Apollo weren't only battling for the title of the GOAT musician. A few millenniums later, they also challenged each other in math (or rather in fast calculations). The task they got to solve is the following:

Let x_1, x_2, …, x_n be the sequence of n non-negative integers. Find this value: $$$∑_{i=1}^n ∑_{j=1}^n ∑_{k=1}^n (x_i   \&   x_j) ⋅ (x_j   |   x_k)$$$

Here \& denotes the [bitwise and,](https://en.wikipedia.org/wiki/Bitwise_operation#AND) and | denotes the [bitwise or.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)

Pan and Apollo could solve this in a few seconds. Can you do it too? For convenience, find the answer modulo 10^9 + 7.

Input

The first line of the input contains a single integer t (1 ≤ t ≤ 1 000) denoting the number of test cases, then t test cases follow.

The first line of each test case consists of a single integer n (1 ≤ n ≤ 5 ⋅ 10^5), the length of the sequence. The second one contains n non-negative integers x_1, x_2, …, x_n (0 ≤ x_i < 2^{60}), elements of the sequence.

The sum of n over all test cases will not exceed 5 ⋅ 10^5.

Output

Print t lines. The i-th line should contain the answer to the i-th text case.

Example

Input


8
2
1 7
3
1 2 4
4
5 5 5 5
5
6 2 2 1 0
1
0
1
1
6
1 12 123 1234 12345 123456
5
536870912 536870911 1152921504606846975 1152921504606846974 1152921504606846973


Output


128
91
1600
505
0
1
502811676
264880351

### Response

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

#define ll long long
#define endl "\n"
#define fastIO ios_base::sync_with_stdio(false)

const ll mod=1e9+7;

int main(){
	fastIO;

	ll t;
	cin>>t;
	while(t--){
		ll n;
		cin>>n;
		ll a[n+1];
		for(ll i=1;i<=n;i++){
			cin>>a[i];
		}
		ll bit[60];
		memset(bit,0,sizeof bit);
		for(ll j=59;j>=0;j--){
			for(ll i=1;i<=n;i++){
				if(a[i]&(1LL<<j)){
					bit[j]++;
				}
			}
		}
		ll ans=0;
		for(ll i=1;i<=n;i++){
			/*ll nd[62];
			memset(nd,0,sizeof nd);
			ll ro[62];
			memset(ro,0,sizeof ro);*/
			ll sum=0;
			for(ll j=59;j>=0;j--){
				ll nd=0;
				if(a[i]&(1LL<<j)){
					nd=(bit[j])%mod;
					ll gun=(1LL<<j)%mod;
					nd=(nd*gun)%mod;
				}
				sum=(sum+nd)%mod;
			}
			for(ll j=59;j>=0;j--){
				ll ro=0;
				if(a[i]&(1LL<<j)){
					ro=n%mod;
					ll gun=(1LL<<j)%mod;
					ro=(ro*gun)%mod;
				}else{
					ro=bit[j]%mod;
					ll gun=(1LL<<j)%mod;
					ro=(ro*gun)%mod;
				}
				ans=(ans+((sum*ro)%mod))%mod;
			}
		}
		cout<<ans<<endl;
	}
}
```