#include <cstdio>
#include <cctype>

template <typename Tp>
void read(Tp &res){
	static char ch; ch = getchar(), res = 0;
	while(!isdigit(ch)) ch = getchar();
	while(isdigit(ch)) res = res * 10 + ch - 48, ch = getchar();
}

typedef long long ll;
const int MAXN = 5e5 + 19, MOD = 1e9 + 7, BIT = 62;

ll a[MAXN];
int n, cnt[BIT], orsum[MAXN], andsum[MAXN];

int main(){
	int T; read(T);
	while(T--){
		read(n);
		for(int i = 0; i < BIT; ++i) cnt[i] = 0;
		for(int i = 1; i <= n; ++i){
			read(a[i]);
			for(int j = 0; j < BIT; ++j)
				if(a[i] & (1ll << j))
					++cnt[j];
		}
		int ans = 0;
		for(int i = 1; i <= n; ++i){
			orsum[i] = 0, andsum[i] = 0;
			for(int j = 0; j < BIT; ++j)
				if(a[i] & (1ll << j))
					orsum[i] = (orsum[i] + (1ll << j) % MOD * n) % MOD, andsum[i] = (andsum[i] + (1ll << j) % MOD * cnt[j]) % MOD;
				else
					orsum[i] = (orsum[i] + (1ll << j) % MOD * cnt[j]) % MOD;
			ans = (ans + (ll)orsum[i] * andsum[i]) % MOD;
		}
		printf("%d\n", (ans + MOD) % MOD);
	}
}
### Prompt

Your challenge is to write 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 <cstdio>
#include <cctype>

template <typename Tp>
void read(Tp &res){
	static char ch; ch = getchar(), res = 0;
	while(!isdigit(ch)) ch = getchar();
	while(isdigit(ch)) res = res * 10 + ch - 48, ch = getchar();
}

typedef long long ll;
const int MAXN = 5e5 + 19, MOD = 1e9 + 7, BIT = 62;

ll a[MAXN];
int n, cnt[BIT], orsum[MAXN], andsum[MAXN];

int main(){
	int T; read(T);
	while(T--){
		read(n);
		for(int i = 0; i < BIT; ++i) cnt[i] = 0;
		for(int i = 1; i <= n; ++i){
			read(a[i]);
			for(int j = 0; j < BIT; ++j)
				if(a[i] & (1ll << j))
					++cnt[j];
		}
		int ans = 0;
		for(int i = 1; i <= n; ++i){
			orsum[i] = 0, andsum[i] = 0;
			for(int j = 0; j < BIT; ++j)
				if(a[i] & (1ll << j))
					orsum[i] = (orsum[i] + (1ll << j) % MOD * n) % MOD, andsum[i] = (andsum[i] + (1ll << j) % MOD * cnt[j]) % MOD;
				else
					orsum[i] = (orsum[i] + (1ll << j) % MOD * cnt[j]) % MOD;
			ans = (ans + (ll)orsum[i] * andsum[i]) % MOD;
		}
		printf("%d\n", (ans + MOD) % MOD);
	}
}
```