#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define vl vector<ll>
#define ml map<ll,ll>
#define sl set<ll>
#define vp vector<pair<ll,ll>>
#define pb push_back

#define asort sort(a,a+n)
#define all(v) v.begin(),v.end()
#define vsort(v) sort(v.begin(),v.end())


#define F first
#define S second
#define mod 1000000007

#define mit map<ll,ll>::iterator
#define for0(j,n) for( i=j;i<n;++i)
#define for1(j,n) for( i=j;i<=n;++i)
#define MA LLONG_MAX
#define MI LLONG_MIN
 
long long power(long long x,ll y)
{
    long long res = 1; 
    x = x % mod;
    while (y > 0) 
    {
        if (y & 1)
            res = (res * x) % mod;
        y = y >> 1; 
        x = (x * x) % mod;
    }
    return res;
}
 
long modInverse(long long n)
{
    return power(n, mod - 2);
}
long long fac[200001];
 
long long ncr(long long n, ll r)
{
    if (n < r)
        return 0;
    if (r == 0)
        return 1;
    long long fac[n + 1];
    fac[0] = 1;
    for (int i = 1; i <= n; i++)
        fac[i] = (fac[i - 1] * i) % mod;
 
    return (fac[n] * modInverse(fac[r]) % mod*modInverse(fac[n - r])% mod)% mod;
}
 void solve()
  {
      ll n;ll i;
      cin>>n;
      ll a[n];
      ll temp,c=0;
      for0(0,n)
        {
            cin>>a[i];
            if(i==0)temp=a[i];
            temp&=a[i];
        }
    for0(0,n)
      {
          if(temp==a[i])
             c++;
      }
      if(c<2)
        {
            cout<<"0";return;
        }
      ll ans=ncr(c,2)%mod;
      ans=(ans%mod*fac[n-2]%mod)%mod*2;
       cout<<ans%mod;return;
  }
 
int main()
{
    ll t=1,i;
    string s1;
    cin>>t;
fac[0] = 1;
    for (int i = 1; i <= 200000; i++)
        fac[i] = (fac[i - 1] * i) % mod;
    while(t--)
    {
        solve();
        cout<<endl;
    }

    return 0;
}
### Prompt

Please provide a CPP coded solution 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 ll long long
#define vl vector<ll>
#define ml map<ll,ll>
#define sl set<ll>
#define vp vector<pair<ll,ll>>
#define pb push_back

#define asort sort(a,a+n)
#define all(v) v.begin(),v.end()
#define vsort(v) sort(v.begin(),v.end())


#define F first
#define S second
#define mod 1000000007

#define mit map<ll,ll>::iterator
#define for0(j,n) for( i=j;i<n;++i)
#define for1(j,n) for( i=j;i<=n;++i)
#define MA LLONG_MAX
#define MI LLONG_MIN
 
long long power(long long x,ll y)
{
    long long res = 1; 
    x = x % mod;
    while (y > 0) 
    {
        if (y & 1)
            res = (res * x) % mod;
        y = y >> 1; 
        x = (x * x) % mod;
    }
    return res;
}
 
long modInverse(long long n)
{
    return power(n, mod - 2);
}
long long fac[200001];
 
long long ncr(long long n, ll r)
{
    if (n < r)
        return 0;
    if (r == 0)
        return 1;
    long long fac[n + 1];
    fac[0] = 1;
    for (int i = 1; i <= n; i++)
        fac[i] = (fac[i - 1] * i) % mod;
 
    return (fac[n] * modInverse(fac[r]) % mod*modInverse(fac[n - r])% mod)% mod;
}
 void solve()
  {
      ll n;ll i;
      cin>>n;
      ll a[n];
      ll temp,c=0;
      for0(0,n)
        {
            cin>>a[i];
            if(i==0)temp=a[i];
            temp&=a[i];
        }
    for0(0,n)
      {
          if(temp==a[i])
             c++;
      }
      if(c<2)
        {
            cout<<"0";return;
        }
      ll ans=ncr(c,2)%mod;
      ans=(ans%mod*fac[n-2]%mod)%mod*2;
       cout<<ans%mod;return;
  }
 
int main()
{
    ll t=1,i;
    string s1;
    cin>>t;
fac[0] = 1;
    for (int i = 1; i <= 200000; i++)
        fac[i] = (fac[i - 1] * i) % mod;
    while(t--)
    {
        solve();
        cout<<endl;
    }

    return 0;
}
```