#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
template<class A, class B>void tostring(pair<A, B>p) { cerr<<"(" + to_string(p.first) + ", " + to_string(p.second) + ")";}
template<class A>void tostring(A v) {int f = 1;string r = "{";for (auto& x : v) {if (!f) {r += ", ";}f = 0;r += to_string(x);}r += "}\n";cerr<<r;}
void _out() { cerr << endl; }template<typename H, typename... T> void _out(H h, T... t) { cerr << ' ' << h; _out(t...); }
void show() { cerr << endl; }template<typename H, typename... T> void show(H h, T... t) { cerr<<h<<' '; show(t...); }
#define see(...) cerr << "(" << #__VA_ARGS__ << "):", _out(__VA_ARGS__)
#define rep(i, l, r) for(int i=(int)(l);i<=(int)(r);i++)
#define per(i, r, l) for(int i=(int)(r);i>=(int)(l);i--)
#define all(x) (x).begin(), (x).end()
#define all1(x) (x).begin()+1, (x).end()
#define pb push_back
#define ub upper_bound
#define lb lower_bound
#define int long long
typedef pair<int, int> pii; typedef vector<int> vi; typedef vector<vi> vvi; typedef vector<pii> vp; typedef vector<vp> vvp;
#define fi first
#define se second
//const int N = 1e5+100;
const int mod=1e9+7;

void solve(int tase) {
	int n;cin >> n;
	vi v(n + 1), f(n + 1);
	map<int, int>cnt;
	
	f[1] = f[0] = 1;
	rep(i, 2, n)f[i] = i * f[i - 1] % mod;
	
	rep(i, 1, n){
		cin >> v[i];
		cnt[v[i]] ++; 
	}
	
	int tmp = v[1];
	rep(i, 2, n){
		tmp &= v[i];
	}
	int ans = 0;
	if(cnt[tmp] < 2)ans = 0;
	else {
		ans += (cnt[tmp] * (cnt[tmp] - 1) % mod) * f[n - 2] % mod;
	}
	cout << ans << endl;
}

signed main() {
	ios::sync_with_stdio(0); cin.tie(0);
#ifndef ONLINE_JUDGE 
	freopen("E:\\in.txt", "r", stdin);
#endif
	int T; cin >> T; rep(i, 1, T) solve(i);
}
### Prompt

Generate a CPP solution to the following problem:

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 <bits/stdc++.h>
using namespace std;
#define endl '\n'
template<class A, class B>void tostring(pair<A, B>p) { cerr<<"(" + to_string(p.first) + ", " + to_string(p.second) + ")";}
template<class A>void tostring(A v) {int f = 1;string r = "{";for (auto& x : v) {if (!f) {r += ", ";}f = 0;r += to_string(x);}r += "}\n";cerr<<r;}
void _out() { cerr << endl; }template<typename H, typename... T> void _out(H h, T... t) { cerr << ' ' << h; _out(t...); }
void show() { cerr << endl; }template<typename H, typename... T> void show(H h, T... t) { cerr<<h<<' '; show(t...); }
#define see(...) cerr << "(" << #__VA_ARGS__ << "):", _out(__VA_ARGS__)
#define rep(i, l, r) for(int i=(int)(l);i<=(int)(r);i++)
#define per(i, r, l) for(int i=(int)(r);i>=(int)(l);i--)
#define all(x) (x).begin(), (x).end()
#define all1(x) (x).begin()+1, (x).end()
#define pb push_back
#define ub upper_bound
#define lb lower_bound
#define int long long
typedef pair<int, int> pii; typedef vector<int> vi; typedef vector<vi> vvi; typedef vector<pii> vp; typedef vector<vp> vvp;
#define fi first
#define se second
//const int N = 1e5+100;
const int mod=1e9+7;

void solve(int tase) {
	int n;cin >> n;
	vi v(n + 1), f(n + 1);
	map<int, int>cnt;
	
	f[1] = f[0] = 1;
	rep(i, 2, n)f[i] = i * f[i - 1] % mod;
	
	rep(i, 1, n){
		cin >> v[i];
		cnt[v[i]] ++; 
	}
	
	int tmp = v[1];
	rep(i, 2, n){
		tmp &= v[i];
	}
	int ans = 0;
	if(cnt[tmp] < 2)ans = 0;
	else {
		ans += (cnt[tmp] * (cnt[tmp] - 1) % mod) * f[n - 2] % mod;
	}
	cout << ans << endl;
}

signed main() {
	ios::sync_with_stdio(0); cin.tie(0);
#ifndef ONLINE_JUDGE 
	freopen("E:\\in.txt", "r", stdin);
#endif
	int T; cin >> T; rep(i, 1, T) solve(i);
}
```