#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int modee=1e9+7;
int Tee,nee;
#define ress ansee=(ansee+pee*qee)%modee;}cout<<ansee<<endl;
int cntee[72];
ll aee[500034];
void solve()
{
     memset(cntee,0,sizeof(cntee));
        cin>>nee;
        for(int i=1; i<=nee; i++)
        {

			scanf("%lld",&aee[i]);
            for(int j=0; j<=60; j++)
            {
                if((1ll<<j)&aee[i])cntee[j]++;
            }
            for(int j=0; j<=60; j++)
            {
                if((1ll<<j)&aee[i])int tf=1;
            }
        }
        ll ansee=0;
        for(int i=1; i<=nee; i++)
        {
            ll pee=0,qee=0;
            for(int j=0; j<=60; j++)
            {
                ll xee=(1ll<<j)%modee;
                if(!((1ll<<j)&aee[i]))
                {
                     pee=(pee+xee*cntee[j])%modee;
                }
                else
                {
                   pee=(pee+xee*nee)%modee;
                    qee=(qee+xee*cntee[j])%modee;
                }


            }ress
}
int main()
{
    cin>>Tee;
    while(Tee--)
    {
       solve();
    }
    // your code goes here
    return 0;
}

### Prompt

Your challenge is to write a CPP solution 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;
typedef long long ll;
const int modee=1e9+7;
int Tee,nee;
#define ress ansee=(ansee+pee*qee)%modee;}cout<<ansee<<endl;
int cntee[72];
ll aee[500034];
void solve()
{
     memset(cntee,0,sizeof(cntee));
        cin>>nee;
        for(int i=1; i<=nee; i++)
        {

			scanf("%lld",&aee[i]);
            for(int j=0; j<=60; j++)
            {
                if((1ll<<j)&aee[i])cntee[j]++;
            }
            for(int j=0; j<=60; j++)
            {
                if((1ll<<j)&aee[i])int tf=1;
            }
        }
        ll ansee=0;
        for(int i=1; i<=nee; i++)
        {
            ll pee=0,qee=0;
            for(int j=0; j<=60; j++)
            {
                ll xee=(1ll<<j)%modee;
                if(!((1ll<<j)&aee[i]))
                {
                     pee=(pee+xee*cntee[j])%modee;
                }
                else
                {
                   pee=(pee+xee*nee)%modee;
                    qee=(qee+xee*cntee[j])%modee;
                }


            }ress
}
int main()
{
    cin>>Tee;
    while(Tee--)
    {
       solve();
    }
    // your code goes here
    return 0;
}

```