#include<bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
const ll Mod = 1000000007;

int t;
int n;
ll x[500500];
ll sum_bit[70];

int main(){
	ios::sync_with_stdio(false);
	cin >> t;
	while(t--){
		cin >> n;
		rep(i, n) cin >> x[i];
		memset(sum_bit, 0, sizeof(sum_bit));
		rep(b, 60) rep(i, n) sum_bit[b] = (sum_bit[b] + (x[i] & (1LL << b))) % Mod;
		ll ans = 0;
		rep(j, n){
			ll sum_and = 0, sum_or = 0;
			rep(b, 60){
				if(x[j] & (1LL << b)) sum_and = (sum_and + sum_bit[b]) % Mod, sum_or = (sum_or + (1LL << b) % Mod * n % Mod) % Mod;
				else sum_or = (sum_or + sum_bit[b]) % Mod;
			}
			ans = (ans + sum_and * sum_or) % Mod;
		}
		cout << ans << endl;
	}
	return 0;
}

### Prompt

Please formulate 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>
#define rep(i, n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
const ll Mod = 1000000007;

int t;
int n;
ll x[500500];
ll sum_bit[70];

int main(){
	ios::sync_with_stdio(false);
	cin >> t;
	while(t--){
		cin >> n;
		rep(i, n) cin >> x[i];
		memset(sum_bit, 0, sizeof(sum_bit));
		rep(b, 60) rep(i, n) sum_bit[b] = (sum_bit[b] + (x[i] & (1LL << b))) % Mod;
		ll ans = 0;
		rep(j, n){
			ll sum_and = 0, sum_or = 0;
			rep(b, 60){
				if(x[j] & (1LL << b)) sum_and = (sum_and + sum_bit[b]) % Mod, sum_or = (sum_or + (1LL << b) % Mod * n % Mod) % Mod;
				else sum_or = (sum_or + sum_bit[b]) % Mod;
			}
			ans = (ans + sum_and * sum_or) % Mod;
		}
		cout << ans << endl;
	}
	return 0;
}

```