
#include <iostream>
#include <stdio.h>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <map>
#include <unordered_map>
#include <math.h>
#include <iomanip>
#include <deque>
#include <set>
#include <queue>
#include <stack>
#include <sstream>

#define _CRT_SECURE_NO_WARNINGS

using namespace std;

#define MAX 200005
const int INF = 1e9;

int fac[MAX] = { 0 };
int mod = 1e9 + 7;

void initFac() {
	fac[0] = 1;
	for (int i = 1; i <= MAX; i++) {
		fac[i] = (1LL * i * fac[i - 1]) % mod;
	}
}

int bigMod(int a, int b, int mod) {
	int result = 1;
	a %= mod;
	while (b != 0) {
		if ((b & 1)) {
			result = (1LL * result * a) % mod;
		}
		b >>= 1;
		a = (1LL * a * a) % mod;
	}
	return result;
}

int inverseMod(int b, int m) {
	int res = bigMod(b, m - 2, m);
	if ((1LL * res * b) % mod == 1) {
		return res;
	}
	return -1;
}

int pombination(int n, int k, int mod) {
	int top = fac[n];
	int bot = fac[n - k];
	return (1LL * top * inverseMod(bot, mod)) % mod;
}

void testCase() {
	int n, a;
	cin >> n;
	unordered_map<int, int> map;
	for (int i = 0; i < n; i++) {
		cin >> a;
		map[a]++;
	}
	unordered_map<int, int>::iterator it, it2;
	int res = 0;
	bool flag;
	for (it = map.begin(); it != map.end(); it++) {
		if (it->second > 1) {
			flag = true;
			for (it2 = map.begin(); it2 != map.end(); it2++) {
				if (((it->first) & (it2->first)) != it->first) {
					flag = false;
					break;
				}
			}
			if (flag) {
				res = (res + 1LL * pombination(it->second, 2, mod) * fac[n - 2]) % mod;
			}
		}
	}
	cout << res << endl;
}

int main() {
	int t;
	cin >> t;
	initFac();
	for (int i = 1; i <= t; i++) {
		testCase();
	}
}
### Prompt

Construct a Cpp code solution to the problem outlined:

A sequence of n non-negative integers (n ≥ 2) a_1, a_2, ..., a_n is called good if for all i from 1 to n-1 the following condition holds true: $$$a_1 \: \& \: a_2 \: \& \: ... \: \& \: a_i = a_{i+1} \: \& \: a_{i+2} \: \& \: ... \: \& \: a_n, where \&$$$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

You are given an array a of size n (n ≥ 2). Find the number of permutations p of numbers ranging from 1 to n, for which the sequence a_{p_1}, a_{p_2}, ... ,a_{p_n} is good. Since this number can be large, output it modulo 10^9+7.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4), denoting the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the size of the array.

The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.

It is guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5.

Output

Output t lines, where the i-th line contains the number of good permutations in the i-th test case modulo 10^9 + 7.

Example

Input


4
3
1 1 1
5
1 2 3 4 5
5
0 2 0 3 0
4
1 3 5 1


Output


6
0
36
4

Note

In the first test case, since all the numbers are equal, whatever permutation we take, the sequence is good. There are a total of 6 permutations possible with numbers from 1 to 3: [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1].

In the second test case, it can be proved that no permutation exists for which the sequence is good.

In the third test case, there are a total of 36 permutations for which the sequence is good. One of them is the permutation [1,5,4,2,3] which results in the sequence s=[0,0,3,2,0]. This is a good sequence because 

  *  s_1 = s_2 \: \& \: s_3 \: \& \: s_4 \: \& \: s_5 = 0, 
  *  s_1 \: \& \: s_2 = s_3 \: \& \: s_4 \: \& \: s_5 = 0, 
  *  s_1 \: \& \: s_2 \: \& \: s_3 = s_4 \: \& \: s_5 = 0, 
  *  s_1 \: \& \: s_2 \: \& \: s_3 \: \& \: s_4 = s_5 = 0. 

### Response

```cpp

#include <iostream>
#include <stdio.h>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <map>
#include <unordered_map>
#include <math.h>
#include <iomanip>
#include <deque>
#include <set>
#include <queue>
#include <stack>
#include <sstream>

#define _CRT_SECURE_NO_WARNINGS

using namespace std;

#define MAX 200005
const int INF = 1e9;

int fac[MAX] = { 0 };
int mod = 1e9 + 7;

void initFac() {
	fac[0] = 1;
	for (int i = 1; i <= MAX; i++) {
		fac[i] = (1LL * i * fac[i - 1]) % mod;
	}
}

int bigMod(int a, int b, int mod) {
	int result = 1;
	a %= mod;
	while (b != 0) {
		if ((b & 1)) {
			result = (1LL * result * a) % mod;
		}
		b >>= 1;
		a = (1LL * a * a) % mod;
	}
	return result;
}

int inverseMod(int b, int m) {
	int res = bigMod(b, m - 2, m);
	if ((1LL * res * b) % mod == 1) {
		return res;
	}
	return -1;
}

int pombination(int n, int k, int mod) {
	int top = fac[n];
	int bot = fac[n - k];
	return (1LL * top * inverseMod(bot, mod)) % mod;
}

void testCase() {
	int n, a;
	cin >> n;
	unordered_map<int, int> map;
	for (int i = 0; i < n; i++) {
		cin >> a;
		map[a]++;
	}
	unordered_map<int, int>::iterator it, it2;
	int res = 0;
	bool flag;
	for (it = map.begin(); it != map.end(); it++) {
		if (it->second > 1) {
			flag = true;
			for (it2 = map.begin(); it2 != map.end(); it2++) {
				if (((it->first) & (it2->first)) != it->first) {
					flag = false;
					break;
				}
			}
			if (flag) {
				res = (res + 1LL * pombination(it->second, 2, mod) * fac[n - 2]) % mod;
			}
		}
	}
	cout << res << endl;
}

int main() {
	int t;
	cin >> t;
	initFac();
	for (int i = 1; i <= t; i++) {
		testCase();
	}
}
```