#include<bits/stdc++.h>
using namespace std;
#define endl "\n"
#define ll long long int
#define pb push_back
#define ff first
#define ss second
#define all(a) (a).begin(),(a).end()
#define vll vector<ll>
#define fl(i,a,b) for(ll i=(ll)a;i<=(ll)b;i++)
#define flr(i,a,b) for(ll i=(ll)a;i>=(ll)b;i--)
#define keep_unique(v) sort(v.begin(), v.end()); v.erase(unique(v.begin(), v.end()), v.end());
#define test() ll te; cin >> te; while(te--)
#define MOD 1000000007
#define fast ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);

const int N=1e6+5;

vll fact(N);

void compute()
{
    fact[0]=1;
    fl(i,1,fact.size()-1) fact[i]=fact[i-1]*i%MOD;
}
//modular exponentiation, n ke jagah n mod(mod-1) se jyada fast hoga
ll modpow(ll x, ll n)
{
	if(n==0) return 1;
	ll u=modpow(x,n/2);
	u=(u*u)%MOD;
	if(n%2==1) u=(u*x)%MOD;
	return u;
}

ll modInverse(ll n)
{
    return modpow(n,MOD-2);
}
ll modmult(ll a, ll b)
{
    return (a*b)%MOD;
}
ll bc(ll n, ll k)
{
	if(n<k) return 0;
    return modmult(modmult(fact[n], modInverse(fact[k])), modInverse(fact[n - k]));
}

int main()
{
	fast;
	compute();
	test()
	{
		ll n; cin>>n;
		vll v(n); fl(i,0,n-1) cin>>v[i];
		ll x=v[0];
		fl(i,0,n-1) x&=v[i];
		ll cnt=0;
		fl(i,0,n-1) if(v[i]==x) cnt++;
		ll ans=bc(cnt,2)*2%MOD;
		ans*=fact[n-2],ans%=MOD;
		cout<<ans<<endl;
	}
}
### Prompt

Develop a solution in Cpp to the problem described below:

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"
#define ll long long int
#define pb push_back
#define ff first
#define ss second
#define all(a) (a).begin(),(a).end()
#define vll vector<ll>
#define fl(i,a,b) for(ll i=(ll)a;i<=(ll)b;i++)
#define flr(i,a,b) for(ll i=(ll)a;i>=(ll)b;i--)
#define keep_unique(v) sort(v.begin(), v.end()); v.erase(unique(v.begin(), v.end()), v.end());
#define test() ll te; cin >> te; while(te--)
#define MOD 1000000007
#define fast ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);

const int N=1e6+5;

vll fact(N);

void compute()
{
    fact[0]=1;
    fl(i,1,fact.size()-1) fact[i]=fact[i-1]*i%MOD;
}
//modular exponentiation, n ke jagah n mod(mod-1) se jyada fast hoga
ll modpow(ll x, ll n)
{
	if(n==0) return 1;
	ll u=modpow(x,n/2);
	u=(u*u)%MOD;
	if(n%2==1) u=(u*x)%MOD;
	return u;
}

ll modInverse(ll n)
{
    return modpow(n,MOD-2);
}
ll modmult(ll a, ll b)
{
    return (a*b)%MOD;
}
ll bc(ll n, ll k)
{
	if(n<k) return 0;
    return modmult(modmult(fact[n], modInverse(fact[k])), modInverse(fact[n - k]));
}

int main()
{
	fast;
	compute();
	test()
	{
		ll n; cin>>n;
		vll v(n); fl(i,0,n-1) cin>>v[i];
		ll x=v[0];
		fl(i,0,n-1) x&=v[i];
		ll cnt=0;
		fl(i,0,n-1) if(v[i]==x) cnt++;
		ll ans=bc(cnt,2)*2%MOD;
		ans*=fact[n-2],ans%=MOD;
		cout<<ans<<endl;
	}
}
```