//#pragma GCC optimize("ofast")
//#pragma GCC optimize("o3")
#include <bits/stdc++.h>
#define int long long
//#define ll long long
#define fr(i,x,y) for(int i=x;i<y;i++)
#define rfr(i,x,y) for(int i=x;i>=y;i--)
using namespace std;
const int N=1e6+10,M=1e2+1,INF=2e9,MOD=1e9+7,L=20,NN=1e9+1;
int mas[N];
int mas2[N];
int k0[61],k1[61];
void solve()
{
	int n;
	cin>>n;
	fr(i,0,n)
	{
		cin>>mas[i];
		mas2[i]=mas[i];
	}
	int ans=0;

	fr(bit,0,61)
	{
		k0[bit]=0;
		k1[bit]=0;
		fr(i,0,n)
		{
			if(mas2[i]%2)k1[bit]++;
			else k0[bit]++;
			mas2[i]/=2;
		}
	}
	fr(j,0,n)
	{
		int sum1=0,sum2=0;
		fr(bit,0,61)
		{
			if(mas[j]%2==1)
			{
				sum1+=k1[bit]*((1LL<<bit)%MOD)%MOD;
				sum2+=n*((1LL<<bit)%MOD)%MOD;
			}
			else
			{
				sum1+=0;
				sum2+=k1[bit]*((1ll<<bit)%MOD)%MOD;
			}
			mas[j]/=2;
			sum1%=MOD;
			sum2%=MOD;
		}
		//cout<<sum1<<' '<<sum2<<'\n';
		ans+=sum1*sum2%MOD;
		ans%=MOD;
	}
	cout<<ans<<'\n';
}
signed main()
{
   //freopen("021.dat","r",stdin);
   //freopen("encryption.out","w",stdout);
   ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
   int t=1;
   cin>>t;
   while(t--)solve();
}
### Prompt

Develop a solution in Cpp to the problem described below:

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
//#pragma GCC optimize("ofast")
//#pragma GCC optimize("o3")
#include <bits/stdc++.h>
#define int long long
//#define ll long long
#define fr(i,x,y) for(int i=x;i<y;i++)
#define rfr(i,x,y) for(int i=x;i>=y;i--)
using namespace std;
const int N=1e6+10,M=1e2+1,INF=2e9,MOD=1e9+7,L=20,NN=1e9+1;
int mas[N];
int mas2[N];
int k0[61],k1[61];
void solve()
{
	int n;
	cin>>n;
	fr(i,0,n)
	{
		cin>>mas[i];
		mas2[i]=mas[i];
	}
	int ans=0;

	fr(bit,0,61)
	{
		k0[bit]=0;
		k1[bit]=0;
		fr(i,0,n)
		{
			if(mas2[i]%2)k1[bit]++;
			else k0[bit]++;
			mas2[i]/=2;
		}
	}
	fr(j,0,n)
	{
		int sum1=0,sum2=0;
		fr(bit,0,61)
		{
			if(mas[j]%2==1)
			{
				sum1+=k1[bit]*((1LL<<bit)%MOD)%MOD;
				sum2+=n*((1LL<<bit)%MOD)%MOD;
			}
			else
			{
				sum1+=0;
				sum2+=k1[bit]*((1ll<<bit)%MOD)%MOD;
			}
			mas[j]/=2;
			sum1%=MOD;
			sum2%=MOD;
		}
		//cout<<sum1<<' '<<sum2<<'\n';
		ans+=sum1*sum2%MOD;
		ans%=MOD;
	}
	cout<<ans<<'\n';
}
signed main()
{
   //freopen("021.dat","r",stdin);
   //freopen("encryption.out","w",stdout);
   ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
   int t=1;
   cin>>t;
   while(t--)solve();
}
```