#include <bits/stdc++.h>

#define int long long

using namespace std;

const int mod = 1e9+7;

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        int a[n];
        int cnt[62];
        for (int i = 0; i < 62; i++) cnt[i] = 0;
        for (int i = 0; i < n; i++) {
            cin >> a[i];
            for (int j = 0; j < 62; j++) if (a[i]&((int)1<<j)) cnt[j]++;
        }
        int res = 0;
        for (int i = 0; i < n; i++) {
            int fsumAnd = 0;
            int fsumOr = 0;
            for (int j = 0; j < 62; j++) {
                if (((int)1<<j)&a[i]) {
                    fsumAnd += cnt[j]*(((int)1<<j)%mod);
                    fsumAnd %= mod;

                    fsumOr += n*(((int)1<<j)%mod);
                    fsumOr %= mod;
                } else {
                    fsumOr += cnt[j] * (((int) 1 << j)%mod);
                    fsumOr %= mod;
                }
            }
            res += fsumAnd*fsumOr;
            res %= mod;
        }
        cout << res << '\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>

#define int long long

using namespace std;

const int mod = 1e9+7;

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        int a[n];
        int cnt[62];
        for (int i = 0; i < 62; i++) cnt[i] = 0;
        for (int i = 0; i < n; i++) {
            cin >> a[i];
            for (int j = 0; j < 62; j++) if (a[i]&((int)1<<j)) cnt[j]++;
        }
        int res = 0;
        for (int i = 0; i < n; i++) {
            int fsumAnd = 0;
            int fsumOr = 0;
            for (int j = 0; j < 62; j++) {
                if (((int)1<<j)&a[i]) {
                    fsumAnd += cnt[j]*(((int)1<<j)%mod);
                    fsumAnd %= mod;

                    fsumOr += n*(((int)1<<j)%mod);
                    fsumOr %= mod;
                } else {
                    fsumOr += cnt[j] * (((int) 1 << j)%mod);
                    fsumOr %= mod;
                }
            }
            res += fsumAnd*fsumOr;
            res %= mod;
        }
        cout << res << '\n';
    }
}
/*

 */
```