#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007LL

ll fact[200005];
ll a[200005];

void calc()
{
    fact[0] = 1LL;
    for(ll i=1;i<=200000;i++)
    {
        fact[i] = ((fact[i-1]%mod) * (i%mod))%mod;
    }
    return ;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    calc();
    int t,n,i,j;

    cin >> t;
    while(t--)
    {
        cin >> n;
        for(i=0;i<n;i++) cin >> a[i];
        sort(a,a + n);

        ll cnt = 1LL;
        for(i=1;i<n;i++)
        {
            if(a[i] == a[0]) cnt++;
            else break;
        }
        if(cnt == 1)
        {
            cout << 0 << endl;
            continue;
        }
        for(i=cnt;i<n;i++)
        {
            if((a[i] & a[0]) != a[0])
            {
                break;
            }
        }
        if(i < n)
        {
            cout << 0 << endl;
            continue;

        }
        ll ans = cnt * (cnt - 1LL);
        ans %= mod;
        ans = ((ans % mod) * (fact[n-2]%mod))%mod;
        cout << ans << endl;
    }
}

### 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 mod 1000000007LL

ll fact[200005];
ll a[200005];

void calc()
{
    fact[0] = 1LL;
    for(ll i=1;i<=200000;i++)
    {
        fact[i] = ((fact[i-1]%mod) * (i%mod))%mod;
    }
    return ;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    calc();
    int t,n,i,j;

    cin >> t;
    while(t--)
    {
        cin >> n;
        for(i=0;i<n;i++) cin >> a[i];
        sort(a,a + n);

        ll cnt = 1LL;
        for(i=1;i<n;i++)
        {
            if(a[i] == a[0]) cnt++;
            else break;
        }
        if(cnt == 1)
        {
            cout << 0 << endl;
            continue;
        }
        for(i=cnt;i<n;i++)
        {
            if((a[i] & a[0]) != a[0])
            {
                break;
            }
        }
        if(i < n)
        {
            cout << 0 << endl;
            continue;

        }
        ll ans = cnt * (cnt - 1LL);
        ans %= mod;
        ans = ((ans % mod) * (fact[n-2]%mod))%mod;
        cout << ans << endl;
    }
}

```