#include <bits/stdc++.h>

using namespace std;

const int P = 1000000000 + 7;

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

	int T;
	cin >> T;

	while (T --) {
		int n;
		cin >> n;

		vector<long long> x(n);
		for (int i = 0; i < n; ++i) {
			cin >> x[i];
		}

		vector<long long> cnt(61);
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < 61; ++j) {
				if (x[i] >> j & 1) {
					cnt[j] ++;
				}
			}
		}

		long long ans = 0;
		for (int j = 0; j < n; ++j) {
			long long sa = 0;
			for (int b = 0; b < 61; ++b) {
				if (x[j] >> b & 1) {
					sa = (sa + (1LL << b) % P * cnt[b] % P) % P;	
				}
			}	
			long long sb = 0;
			for (int b = 0; b < 61; ++b) {
				if (x[j] >> b & 1) {
					sb = (sb + (1LL << b) % P * n % P) % P;	
				} else {
					sb = (sb + (1LL << b) % P * cnt[b] % P) % P;
				}
			}
			ans = (ans + sa * sb) % P;
		}

		cout << ans << '\n';
	}

	return 0;
}
### Prompt

Your task is to create 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;

const int P = 1000000000 + 7;

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

	int T;
	cin >> T;

	while (T --) {
		int n;
		cin >> n;

		vector<long long> x(n);
		for (int i = 0; i < n; ++i) {
			cin >> x[i];
		}

		vector<long long> cnt(61);
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < 61; ++j) {
				if (x[i] >> j & 1) {
					cnt[j] ++;
				}
			}
		}

		long long ans = 0;
		for (int j = 0; j < n; ++j) {
			long long sa = 0;
			for (int b = 0; b < 61; ++b) {
				if (x[j] >> b & 1) {
					sa = (sa + (1LL << b) % P * cnt[b] % P) % P;	
				}
			}	
			long long sb = 0;
			for (int b = 0; b < 61; ++b) {
				if (x[j] >> b & 1) {
					sb = (sb + (1LL << b) % P * n % P) % P;	
				} else {
					sb = (sb + (1LL << b) % P * cnt[b] % P) % P;
				}
			}
			ans = (ans + sa * sb) % P;
		}

		cout << ans << '\n';
	}

	return 0;
}
```