#include<bits/stdc++.h>
using namespace std;
typedef long long int llint;
typedef pair<int, int> pii;
const int MAX = 5e5 + 10, MOD = 1e9 + 7;
int t, n, sum, sol;
llint x[MAX];
int koliko[60];
llint pot[61];
int add(int x, int y){
	return (x + y) % MOD;
}
int mul(int x, int y){
	llint ret = (llint)x * y;
	return ret % MOD;
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);cout.tie(NULL);
	llint y = 2;pot[0] = 1;
	for(int i = 1;i <= 60; i++){
		pot[i] = pot[i - 1] * y;
	}
	cin >> t;
	while(t--){
		cin >> n;
		for(int i = 0;i < n; i++){
			cin >> x[i];
		}
		for(int j = 0;j < 60; j++){
			int kol = 0;
			for(int i = 0;i < n; i++){
				if(x[i] & pot[j]) kol++;
			}
			koliko[j] = kol;
		}
		sum = 0;
		for(int j = 0;j < n; j++){
			int k1 = 0, k2 = 0;
			for(int i = 0;i < 60; i++){
				if(x[j] & pot[i]){
					k1 = add(k1, mul(koliko[i], pot[i] % MOD));
					k2 = add(k2, mul(n, pot[i] % MOD));
				}
				else{
					k2 = add(k2, mul(koliko[i], pot[i] % MOD));
				}
			}
			sum = add(sum, mul(k1, k2));
		}
		cout << sum << "\n";
	}
	return 0;
}

### Prompt

Generate 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;
typedef long long int llint;
typedef pair<int, int> pii;
const int MAX = 5e5 + 10, MOD = 1e9 + 7;
int t, n, sum, sol;
llint x[MAX];
int koliko[60];
llint pot[61];
int add(int x, int y){
	return (x + y) % MOD;
}
int mul(int x, int y){
	llint ret = (llint)x * y;
	return ret % MOD;
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);cout.tie(NULL);
	llint y = 2;pot[0] = 1;
	for(int i = 1;i <= 60; i++){
		pot[i] = pot[i - 1] * y;
	}
	cin >> t;
	while(t--){
		cin >> n;
		for(int i = 0;i < n; i++){
			cin >> x[i];
		}
		for(int j = 0;j < 60; j++){
			int kol = 0;
			for(int i = 0;i < n; i++){
				if(x[i] & pot[j]) kol++;
			}
			koliko[j] = kol;
		}
		sum = 0;
		for(int j = 0;j < n; j++){
			int k1 = 0, k2 = 0;
			for(int i = 0;i < 60; i++){
				if(x[j] & pot[i]){
					k1 = add(k1, mul(koliko[i], pot[i] % MOD));
					k2 = add(k2, mul(n, pot[i] % MOD));
				}
				else{
					k2 = add(k2, mul(koliko[i], pot[i] % MOD));
				}
			}
			sum = add(sum, mul(k1, k2));
		}
		cout << sum << "\n";
	}
	return 0;
}

```