#include<bits/stdc++.h>

using namespace std;

#define MOD 1000000007
typedef long long int ll;

int main(){

	#ifndef ONLINE_JUDGE
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	#endif

	ios_base::sync_with_stdio(false);
	cin.tie(0);
        
	ll t;
	cin>>t;
	while(t--){
		ll n;
		cin>>n;
		vector<ll>a(n,0);
		vector<ll>cnt(62,0);
		for(ll i=0;i<n;i++){
			cin>>a[i];
			for(ll j=0;j<=61;j++){
				if((a[i]&(1LL<<j))){
					cnt[j]++;
				}
			}
		}
		ll ans=0;
		for(ll i=0;i<n;i++){
			ll an=0;
			ll oo=0;
			for(ll j=0;j<=61;j++){
				if((a[i]&(1LL<<j))){
					an=(an+(((cnt[j])*((1LL<<j)%MOD))%MOD))%MOD;
					oo=(oo+((n*((1LL<<j)%MOD))%MOD))%MOD;
				}
				else{
					oo=(oo+(((cnt[j])*((1LL<<j)%MOD))%MOD))%MOD;
				}
			}
			ans=(ans+((an*oo)%MOD))%MOD;
		}
		cout<<ans<<endl;


	}

	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 MOD 1000000007
typedef long long int ll;

int main(){

	#ifndef ONLINE_JUDGE
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
	#endif

	ios_base::sync_with_stdio(false);
	cin.tie(0);
        
	ll t;
	cin>>t;
	while(t--){
		ll n;
		cin>>n;
		vector<ll>a(n,0);
		vector<ll>cnt(62,0);
		for(ll i=0;i<n;i++){
			cin>>a[i];
			for(ll j=0;j<=61;j++){
				if((a[i]&(1LL<<j))){
					cnt[j]++;
				}
			}
		}
		ll ans=0;
		for(ll i=0;i<n;i++){
			ll an=0;
			ll oo=0;
			for(ll j=0;j<=61;j++){
				if((a[i]&(1LL<<j))){
					an=(an+(((cnt[j])*((1LL<<j)%MOD))%MOD))%MOD;
					oo=(oo+((n*((1LL<<j)%MOD))%MOD))%MOD;
				}
				else{
					oo=(oo+(((cnt[j])*((1LL<<j)%MOD))%MOD))%MOD;
				}
			}
			ans=(ans+((an*oo)%MOD))%MOD;
		}
		cout<<ans<<endl;


	}

	return 0;
}
```