#include<iostream>

using namespace std;


#define ll long long int
int mod = 1e9+7;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        int ar[n+1];
        int mn =mod;
        for(int i= 0;i<n;i++)
        {
            cin>>ar[i];
            mn  = min(mn,ar[i]);
        }
        int cnt =0;
        bool flag = 0;
        for(int i = 0;i<n;i++)
        {
            if((mn & ar[i])  != mn)
            {
                cout<<0<<'\n';


                flag = 1;
                break;
            }
            if(mn == ar[i])
            cnt++;
        }
if(flag)
    continue;

        ll ans = 0;

        ans = (1ll * cnt*(cnt-1))%mod;

        ll fact = 1;
        for(int i =1;i<=(n-2);i++)
        {
            fact = (fact * i)%mod;
        }

        ans = (ans *fact)%mod;

        cout<<ans<<'\n';

    }

}

### 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<iostream>

using namespace std;


#define ll long long int
int mod = 1e9+7;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        int ar[n+1];
        int mn =mod;
        for(int i= 0;i<n;i++)
        {
            cin>>ar[i];
            mn  = min(mn,ar[i]);
        }
        int cnt =0;
        bool flag = 0;
        for(int i = 0;i<n;i++)
        {
            if((mn & ar[i])  != mn)
            {
                cout<<0<<'\n';


                flag = 1;
                break;
            }
            if(mn == ar[i])
            cnt++;
        }
if(flag)
    continue;

        ll ans = 0;

        ans = (1ll * cnt*(cnt-1))%mod;

        ll fact = 1;
        for(int i =1;i<=(n-2);i++)
        {
            fact = (fact * i)%mod;
        }

        ans = (ans *fact)%mod;

        cout<<ans<<'\n';

    }

}

```