#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>

using namespace std;

const int maxn = 500007;
long long x[maxn] = {0};
const int smax = 64;
long long a[smax] = {0};
long long stm[smax] = {0};
long long mod = 1000000007;

int main()
{
    long long s = 1;
    for(int i = 0; i < smax; ++i)
    {
        stm[i] = s;
        s *= 2;
        s = s%mod;
    }
    int ttt;
    cin >> ttt;
    for(int iii = 0; iii < ttt; ++iii)
    {
        int n;
        scanf("%d",&n);
        long long t;
        int ch = 0;
        for(int i = 0; i < smax; ++i)
            a[i] = 0;
        for(int i = 0; i < n; ++i)
        {
            scanf("%I64d", &x[i]);
            t = x[i];
            ch = 0;
            while(t > 0)
            {
                if(t%2 == 1)
                    a[ch]++;
                t /= 2;
                ch++;
            }
        }
        long long res = 0;
        for(int j = 0; j < n; ++j)
        {
            t = x[j];
            long long l = 0;
            long long r = 0;
            for(int i = 0; i < smax; ++i)
            {
                if(t%2 == 1)
                {
                    l += a[i]*stm[i];
                    l = l % mod;
                    r += (long long)(n)*stm[i];
                    r = r % mod;
                }
                else
                {
                    r += a[i]*stm[i];
                    r = r % mod;
                }
                t /= 2;
            }
            res += l*r;
            res = res % mod;
        }
        cout << res << '\n';
    }

    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 <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>

using namespace std;

const int maxn = 500007;
long long x[maxn] = {0};
const int smax = 64;
long long a[smax] = {0};
long long stm[smax] = {0};
long long mod = 1000000007;

int main()
{
    long long s = 1;
    for(int i = 0; i < smax; ++i)
    {
        stm[i] = s;
        s *= 2;
        s = s%mod;
    }
    int ttt;
    cin >> ttt;
    for(int iii = 0; iii < ttt; ++iii)
    {
        int n;
        scanf("%d",&n);
        long long t;
        int ch = 0;
        for(int i = 0; i < smax; ++i)
            a[i] = 0;
        for(int i = 0; i < n; ++i)
        {
            scanf("%I64d", &x[i]);
            t = x[i];
            ch = 0;
            while(t > 0)
            {
                if(t%2 == 1)
                    a[ch]++;
                t /= 2;
                ch++;
            }
        }
        long long res = 0;
        for(int j = 0; j < n; ++j)
        {
            t = x[j];
            long long l = 0;
            long long r = 0;
            for(int i = 0; i < smax; ++i)
            {
                if(t%2 == 1)
                {
                    l += a[i]*stm[i];
                    l = l % mod;
                    r += (long long)(n)*stm[i];
                    r = r % mod;
                }
                else
                {
                    r += a[i]*stm[i];
                    r = r % mod;
                }
                t /= 2;
            }
            res += l*r;
            res = res % mod;
        }
        cout << res << '\n';
    }

    return 0;
}

```