// Overidea
#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=l;i<=r;++i)
#define repdo(i,l,r) for(int i=l;i>=r;--i)
#define ll long long
using namespace std;
//----------------------------------------------------
const int nsz=5e5+50;
const ll nmod=1e9+7;
ll t,n,li[nsz];
int cnt[100];
ll v1[nsz],v2[nsz];//or and
int main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);

	cin>>t;
	rep(cs,1,t){
		cin>>n;
		rep(i,1,n)cin>>li[i];
		rep(i,0,61)cnt[i]=0;
		rep(i,1,n)v1[i]=v2[i]=0;
		rep(b,0,61)rep(i,1,n)cnt[b]+=((li[i]>>b)&1);
		rep(i,1,n){
			rep(b,0,61){
				if((li[i]>>b)&1)v1[i]=(v1[i]+(1ll<<b)%nmod*n)%nmod;
				else v1[i]=(v1[i]+(1ll<<b)%nmod*cnt[b])%nmod;
				
				if((li[i]>>b)&1)v2[i]=(v2[i]+(1ll<<b)%nmod*(cnt[b]))%nmod;
			}
		}
		ll res=0;
		rep(i,1,n)res=(res+v1[i]*v2[i])%nmod;
		cout<<res<<'\n';
	}

	return 0;
}




### Prompt

In Cpp, your task is to solve 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
// Overidea
#include<bits/stdc++.h>
#define rep(i,l,r) for(int i=l;i<=r;++i)
#define repdo(i,l,r) for(int i=l;i>=r;--i)
#define ll long long
using namespace std;
//----------------------------------------------------
const int nsz=5e5+50;
const ll nmod=1e9+7;
ll t,n,li[nsz];
int cnt[100];
ll v1[nsz],v2[nsz];//or and
int main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);

	cin>>t;
	rep(cs,1,t){
		cin>>n;
		rep(i,1,n)cin>>li[i];
		rep(i,0,61)cnt[i]=0;
		rep(i,1,n)v1[i]=v2[i]=0;
		rep(b,0,61)rep(i,1,n)cnt[b]+=((li[i]>>b)&1);
		rep(i,1,n){
			rep(b,0,61){
				if((li[i]>>b)&1)v1[i]=(v1[i]+(1ll<<b)%nmod*n)%nmod;
				else v1[i]=(v1[i]+(1ll<<b)%nmod*cnt[b])%nmod;
				
				if((li[i]>>b)&1)v2[i]=(v2[i]+(1ll<<b)%nmod*(cnt[b]))%nmod;
			}
		}
		ll res=0;
		rep(i,1,n)res=(res+v1[i]*v2[i])%nmod;
		cout<<res<<'\n';
	}

	return 0;
}




```