#include <bits/stdc++.h>

using namespace std;

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    long long mod = 1e9 + 7;
    int t;
    cin >> t;
    while (t--) {
        int n;
        long long ans = 0;
        cin >> n;
        vector<long long> a(n);
        vector<long long> bits(60, 0);
        for (int q = 0; q < n; q++) {
            cin >> a[q];
            long long curr = a[q];
            for (int w = 0; w < 60; w++) {
                bits[w] += curr % 2;
                curr = curr >> 1;
            }
        }
        for (int q = 0; q < n; q++) {
            long long first = 0, second = 0;
            long long curr2 = 1;
            long long curr = a[q];
            for (int w = 0; w < 60; w++) {
                long long bit = curr % 2;
                curr = curr >> 1;
                if (bit == 1) {
                    first = (first + curr2 * bits[w]) % mod;
                    second = (second + curr2 * n) % mod;
                }
                else {
                    second = (second + curr2 * bits[w]) % mod;
                }
                curr2 = (curr2 * 2) % mod;
            }
            ans = (ans + first * second) % mod;
        }
        cout << ans << "\n";
    }
    return 0;
}

### 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;

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    long long mod = 1e9 + 7;
    int t;
    cin >> t;
    while (t--) {
        int n;
        long long ans = 0;
        cin >> n;
        vector<long long> a(n);
        vector<long long> bits(60, 0);
        for (int q = 0; q < n; q++) {
            cin >> a[q];
            long long curr = a[q];
            for (int w = 0; w < 60; w++) {
                bits[w] += curr % 2;
                curr = curr >> 1;
            }
        }
        for (int q = 0; q < n; q++) {
            long long first = 0, second = 0;
            long long curr2 = 1;
            long long curr = a[q];
            for (int w = 0; w < 60; w++) {
                long long bit = curr % 2;
                curr = curr >> 1;
                if (bit == 1) {
                    first = (first + curr2 * bits[w]) % mod;
                    second = (second + curr2 * n) % mod;
                }
                else {
                    second = (second + curr2 * bits[w]) % mod;
                }
                curr2 = (curr2 * 2) % mod;
            }
            ans = (ans + first * second) % mod;
        }
        cout << ans << "\n";
    }
    return 0;
}

```