#include<bits/stdc++.h>
using namespace std;
long long int x[500010];
int mod = 1e9 + 7;
int cnt[61];
long long int ok[61];
long long int okn[61];
void solve() {
	int n;
	cin>>n;
	memset(cnt,0,sizeof cnt);
	memset(ok,0,sizeof ok);
	memset(okn,0,sizeof okn);
	for(int i = 0; i < n; i++) {
		cin>>x[i];
		for(int j = 59; j >= 0; j--) {
			if(x[i] & (1LL << j)) {
				cnt[j]++;
			}
		}
	}
	okn[0] = n;
	for(int j = 0; j <= 59; j++) {
		if(j > 0)
			okn[j] = okn[j - 1] * 2 % mod;
		
		ok[j] = cnt[j];
		if(j >= 30) {
			ok[j] = (ok[j] << 30) % mod;
			ok[j] = (ok[j] << (j - 30)) % mod;
		}
		else {
			ok[j] = (ok[j] << j) % mod;
		}
	}
	int ans = 0;
	for(int i = 0; i < n; i++) {
		long long int sumand = 0;
		long long int sumor = 0;
		for(int j = 59; j >= 0; j--) {
			if(x[i] & (1LL << j)) {
				sumand = (sumand + ok[j]);
				sumor = (sumor + okn[j]);
			} else {
				sumor = (sumor + ok[j]);
			}
		}
		sumand %= mod;
		sumor %= mod;
		ans = (ans + (sumand * sumor)) % mod;
	}
	cout<<ans<<"\n";
}
int main() {
#ifndef ONLINE_JUDGE
	freopen("test.inp", "r", stdin);
	freopen("test.out", "w", stdout);
#endif
	int t;
	cin>>t;
	while(t--)
		solve();
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;
long long int x[500010];
int mod = 1e9 + 7;
int cnt[61];
long long int ok[61];
long long int okn[61];
void solve() {
	int n;
	cin>>n;
	memset(cnt,0,sizeof cnt);
	memset(ok,0,sizeof ok);
	memset(okn,0,sizeof okn);
	for(int i = 0; i < n; i++) {
		cin>>x[i];
		for(int j = 59; j >= 0; j--) {
			if(x[i] & (1LL << j)) {
				cnt[j]++;
			}
		}
	}
	okn[0] = n;
	for(int j = 0; j <= 59; j++) {
		if(j > 0)
			okn[j] = okn[j - 1] * 2 % mod;
		
		ok[j] = cnt[j];
		if(j >= 30) {
			ok[j] = (ok[j] << 30) % mod;
			ok[j] = (ok[j] << (j - 30)) % mod;
		}
		else {
			ok[j] = (ok[j] << j) % mod;
		}
	}
	int ans = 0;
	for(int i = 0; i < n; i++) {
		long long int sumand = 0;
		long long int sumor = 0;
		for(int j = 59; j >= 0; j--) {
			if(x[i] & (1LL << j)) {
				sumand = (sumand + ok[j]);
				sumor = (sumor + okn[j]);
			} else {
				sumor = (sumor + ok[j]);
			}
		}
		sumand %= mod;
		sumor %= mod;
		ans = (ans + (sumand * sumor)) % mod;
	}
	cout<<ans<<"\n";
}
int main() {
#ifndef ONLINE_JUDGE
	freopen("test.inp", "r", stdin);
	freopen("test.out", "w", stdout);
#endif
	int t;
	cin>>t;
	while(t--)
		solve();
}
```