// AUTHOR:-  Harsh Thete
//   IIIT-Allahabad
#include <bits/stdc++.h>
using namespace std;

#define ll long long

#define fl(e,a,b) for(ll e=a;e<b;e++)

#define fst ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define T 	int T;cin>>T;while(T--)

ll m=1e9+7;ll p[60];
void my_ans_is()
{
	ll n;cin>>n;ll ans=0;vector<ll>a(n);vector<ll> b(60,0);
	
	fl(i,0,n)
	{
		cin>>a[i];
		fl(j,0,60)
		{
			if((1LL<<j)&a[i]) b[j]++;
		}
	}
	fl(i,0,n)
	{
		ll x=0,y=0;
		fl(j,0,60)
		{
			if((1LL<<j)&a[i])
				{x=(x%m+(b[j]%m*p[j]%m)%m)%m; y=(y%m+((n%m)*(p[j]%m))%m)%m;}
			else 
				y=(y%m+(b[j]%m*p[j]%m)%m)%m;
		}
		ans=(ans%m+(x%m*y%m)%m)%m;
	}
	cout<<ans<<"\n";
}

//cin.getline(s,1000) to get string cointaining spaces where s->char s[1000] to find exact len n=strlen(s);

int main() 
{         // Euoiler tot phi(n)=cnt of numbers less than n (from 1to n) whose gcd with n is 1
	     //  for tot cnt of gcd=1 from x to n-> cnt= phi(n/gcd(x,n))
	fst; 
    
	   p[0]=1;                            //////////////__gcd(x,y)=__gcd(x-y,y)
	fl(i,1,60)
	{
	   p[i]=(p[i-1]*2)%m;
	}
	T
	{      
         my_ans_is();
// Note:-Check bounds in arrays, check limit of int (signed integer overflow)
        }
   return 0;
}
### 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
// AUTHOR:-  Harsh Thete
//   IIIT-Allahabad
#include <bits/stdc++.h>
using namespace std;

#define ll long long

#define fl(e,a,b) for(ll e=a;e<b;e++)

#define fst ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define T 	int T;cin>>T;while(T--)

ll m=1e9+7;ll p[60];
void my_ans_is()
{
	ll n;cin>>n;ll ans=0;vector<ll>a(n);vector<ll> b(60,0);
	
	fl(i,0,n)
	{
		cin>>a[i];
		fl(j,0,60)
		{
			if((1LL<<j)&a[i]) b[j]++;
		}
	}
	fl(i,0,n)
	{
		ll x=0,y=0;
		fl(j,0,60)
		{
			if((1LL<<j)&a[i])
				{x=(x%m+(b[j]%m*p[j]%m)%m)%m; y=(y%m+((n%m)*(p[j]%m))%m)%m;}
			else 
				y=(y%m+(b[j]%m*p[j]%m)%m)%m;
		}
		ans=(ans%m+(x%m*y%m)%m)%m;
	}
	cout<<ans<<"\n";
}

//cin.getline(s,1000) to get string cointaining spaces where s->char s[1000] to find exact len n=strlen(s);

int main() 
{         // Euoiler tot phi(n)=cnt of numbers less than n (from 1to n) whose gcd with n is 1
	     //  for tot cnt of gcd=1 from x to n-> cnt= phi(n/gcd(x,n))
	fst; 
    
	   p[0]=1;                            //////////////__gcd(x,y)=__gcd(x-y,y)
	fl(i,1,60)
	{
	   p[i]=(p[i-1]*2)%m;
	}
	T
	{      
         my_ans_is();
// Note:-Check bounds in arrays, check limit of int (signed integer overflow)
        }
   return 0;
}
```