#include <bits/stdc++.h>

#define pb push_back
#define int long long 
#define debug(n) cerr << #n << " = " << n << '\n';

using namespace std;

const int MaxN = 1e6 + 10;
const int Mod = 1e9 + 7;
const int inf = 1e18;
int cnt[100] , a[MaxN];

void solve() {
	int n;
	cin >> n;
	for(int i = 0; i < n; i++) {
		cin >> a[i];
		for(int j = 0; j <= 60; j++) 
			if(a[i] & (1LL << j)) cnt[j]++;
	}
	int ans = 0;
	for(int i = 0; i < n; i++) {
		int x = 0 , y = 0;
		for(int j = 0; j <= 60; j++) {
			int p = (1LL << j) % Mod;
			if(a[i] & (1LL << j)) {
				x = (x + p * cnt[j]) % Mod;
				y = (y + p * n) % Mod;
			} else y = (y + p * cnt[j]) % Mod;
		}
		ans = (ans + x * y) % Mod;
	}
	cout << ans << '\n';
	fill(cnt , cnt + 100 , 0);
}

int32_t main() {
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int t = 1;
	cin >> t;
	while(t--) 
		solve();
	return 0;
} 
### Prompt

Create a solution in cpp for 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>

#define pb push_back
#define int long long 
#define debug(n) cerr << #n << " = " << n << '\n';

using namespace std;

const int MaxN = 1e6 + 10;
const int Mod = 1e9 + 7;
const int inf = 1e18;
int cnt[100] , a[MaxN];

void solve() {
	int n;
	cin >> n;
	for(int i = 0; i < n; i++) {
		cin >> a[i];
		for(int j = 0; j <= 60; j++) 
			if(a[i] & (1LL << j)) cnt[j]++;
	}
	int ans = 0;
	for(int i = 0; i < n; i++) {
		int x = 0 , y = 0;
		for(int j = 0; j <= 60; j++) {
			int p = (1LL << j) % Mod;
			if(a[i] & (1LL << j)) {
				x = (x + p * cnt[j]) % Mod;
				y = (y + p * n) % Mod;
			} else y = (y + p * cnt[j]) % Mod;
		}
		ans = (ans + x * y) % Mod;
	}
	cout << ans << '\n';
	fill(cnt , cnt + 100 , 0);
}

int32_t main() {
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int t = 1;
	cin >> t;
	while(t--) 
		solve();
	return 0;
} 
```