#include <bits/stdc++.h>

using namespace std;

#define int ll
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<double> vd;
typedef vector<bool> vb;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<pi> vp;
typedef vector<pl> vpl;

const int mod = 1e9+7;

void f(int& x){
    if (x >= mod)
        x %= mod;
}

int g(int x){
    f(x);
    return x;
}

signed main()
{
    ios_base::sync_with_stdio(false); 
    cin.tie(nullptr); 
    cout.tie(nullptr); 
    cerr.tie(nullptr);    

    int t;
    cin >> t;

    for (int e = 0; e < t; ++e){
        int n;
        cin >> n;

        vi a(n), cnt(60);
        for (auto& i : a){
            cin >> i;

            for (int j = 0; j < 60; ++j)
                if ((1ll << j) & i)
                    ++cnt[j];
        }

        vi suma(n, 0);
        for (int i = 0; i < n; ++i){
            for (int j = 0; j < 60; ++j){
                if (a[i]&(1ll<<j)){
                    suma[i] += g(cnt[j] * g(1ll << j));
                    f(suma[i]);
                }
            }
        }

        int sol = 0;
        //for (int i = 0; i < n; ++i){
            for (int j = 0; j < n; ++j){
                int x = suma[j];
                f(x);

                for (int k = 0; k < 60; ++k){
                    int y = 1ll << k;
                    f(y);
                    if ((1ll << k) & a[j]){
                        sol += g(n*g(y * x));
                        f(sol);
                    } else {
                        sol += g(g(x * cnt[k])*y);
                        f(sol);
                    }
                }

//                cout << sol << '\n';
            }
    //    }

        cout << sol << '\n';
    }
}
### 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
#include <bits/stdc++.h>

using namespace std;

#define int ll
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<double> vd;
typedef vector<bool> vb;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<pi> vp;
typedef vector<pl> vpl;

const int mod = 1e9+7;

void f(int& x){
    if (x >= mod)
        x %= mod;
}

int g(int x){
    f(x);
    return x;
}

signed main()
{
    ios_base::sync_with_stdio(false); 
    cin.tie(nullptr); 
    cout.tie(nullptr); 
    cerr.tie(nullptr);    

    int t;
    cin >> t;

    for (int e = 0; e < t; ++e){
        int n;
        cin >> n;

        vi a(n), cnt(60);
        for (auto& i : a){
            cin >> i;

            for (int j = 0; j < 60; ++j)
                if ((1ll << j) & i)
                    ++cnt[j];
        }

        vi suma(n, 0);
        for (int i = 0; i < n; ++i){
            for (int j = 0; j < 60; ++j){
                if (a[i]&(1ll<<j)){
                    suma[i] += g(cnt[j] * g(1ll << j));
                    f(suma[i]);
                }
            }
        }

        int sol = 0;
        //for (int i = 0; i < n; ++i){
            for (int j = 0; j < n; ++j){
                int x = suma[j];
                f(x);

                for (int k = 0; k < 60; ++k){
                    int y = 1ll << k;
                    f(y);
                    if ((1ll << k) & a[j]){
                        sol += g(n*g(y * x));
                        f(sol);
                    } else {
                        sol += g(g(x * cnt[k])*y);
                        f(sol);
                    }
                }

//                cout << sol << '\n';
            }
    //    }

        cout << sol << '\n';
    }
}
```