
#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstring>
#include <chrono>
#include <complex>
#define endl "\n"
#define ll long long int
#define vi vector<int>
#define vll vector<ll>
#define vvi vector < vi >
#define pii pair<int,int>
#define pll pair<long long, long long>
#define mod 1000000007
#define inf 1000000000000000001;
#define all(c) c.begin(),c.end()
#define mp(x,y) make_pair(x,y)
#define mem(a,val) memset(a,val,sizeof(a))
#define eb emplace_back
#define f first
#define s second
#define pb push_back
#define pp pop_back

using namespace std;
ll fact[200005];
void solve()
{
	ll n;cin>>n;
	ll a[n];
	for(ll i=0;i<n;i++)cin>>a[i];
	ll t=a[0];
	for(ll i=0;i<n;i++)
	{
		t=t&a[i];
	}
	ll cnt=0;
	for(ll i=0;i<n;i++)
	{
		if(a[i]==t)cnt++;
	}
	if(cnt<2)cout<<0<<endl;
	else
	{
		cout<<(cnt%mod*(cnt-1)%mod*fact[n-2])%mod<<endl;
	}
}

int main()
{
  std::ios::sync_with_stdio(false);
  ll T=1;
  cin>>T;
	fact[0]=1;
	fact[1]=1;
	for(int i=2;i<200005;i++)
	fact[i]=(i%mod*fact[i-1]%mod)%mod;
  // cin.ignore(); must be there when using getline(cin, s)
  while(T--)
  {
    solve();
  }


  return 0;
}

### Prompt

Please create a solution in Cpp 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>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstring>
#include <chrono>
#include <complex>
#define endl "\n"
#define ll long long int
#define vi vector<int>
#define vll vector<ll>
#define vvi vector < vi >
#define pii pair<int,int>
#define pll pair<long long, long long>
#define mod 1000000007
#define inf 1000000000000000001;
#define all(c) c.begin(),c.end()
#define mp(x,y) make_pair(x,y)
#define mem(a,val) memset(a,val,sizeof(a))
#define eb emplace_back
#define f first
#define s second
#define pb push_back
#define pp pop_back

using namespace std;
ll fact[200005];
void solve()
{
	ll n;cin>>n;
	ll a[n];
	for(ll i=0;i<n;i++)cin>>a[i];
	ll t=a[0];
	for(ll i=0;i<n;i++)
	{
		t=t&a[i];
	}
	ll cnt=0;
	for(ll i=0;i<n;i++)
	{
		if(a[i]==t)cnt++;
	}
	if(cnt<2)cout<<0<<endl;
	else
	{
		cout<<(cnt%mod*(cnt-1)%mod*fact[n-2])%mod<<endl;
	}
}

int main()
{
  std::ios::sync_with_stdio(false);
  ll T=1;
  cin>>T;
	fact[0]=1;
	fact[1]=1;
	for(int i=2;i<200005;i++)
	fact[i]=(i%mod*fact[i-1]%mod)%mod;
  // cin.ignore(); must be there when using getline(cin, s)
  while(T--)
  {
    solve();
  }


  return 0;
}

```