#include <iostream>
#include <vector>
#include <algorithm>
#include <map>

typedef long long ll;

ll mod = 1e9 + 7;

ll powmod(ll a, ll n)
{
    if (n == 0)
        return 1;
    if (n == 1)
        return a % mod;

    ll b = powmod(a, n / 2);
    b = (b * b) % mod;
    if (n & 1)
    {
        b = (b * a) % mod;
    }
    return b % mod;
}

std::vector<ll> fact(202020);

int main(void)
{
    std::cin.tie(0);
    std::ios_base::sync_with_stdio(false);

    fact[0] = 1;
    fact[1] = 1;
    for (int i = 2; i < 202020; i++)
    {
        fact[i] = (fact[i - 1] * i) % mod;
    }

    int T;
    std::cin >> T;
    while (T--)
    {
        int N;
        std::cin >> N;
        std::vector<ll> v(N);
        std::map<ll, ll> mp;
        ll AND = __LONG_LONG_MAX__;
        ll x = 1;

        for (int i = 0; i < N; i++)
        {
            std::cin >> v[i];
            AND &= v[i];
            mp[v[i]]++;
            x *= mp[v[i]];
            x %= mod;
        }

        ll res = 0;
        ll cnt = 0;

        for (int i = 0; i < N; i++)
        {
            if ((AND == v[i]))
            {
                cnt++;
            }
        }

        if (cnt >= 2)
        {
            res = ((cnt - 1) * cnt) % mod;
            res = (res * fact[N - 2]) % mod;
        }

        std::cout << res % mod << "\n";
    }

    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 <iostream>
#include <vector>
#include <algorithm>
#include <map>

typedef long long ll;

ll mod = 1e9 + 7;

ll powmod(ll a, ll n)
{
    if (n == 0)
        return 1;
    if (n == 1)
        return a % mod;

    ll b = powmod(a, n / 2);
    b = (b * b) % mod;
    if (n & 1)
    {
        b = (b * a) % mod;
    }
    return b % mod;
}

std::vector<ll> fact(202020);

int main(void)
{
    std::cin.tie(0);
    std::ios_base::sync_with_stdio(false);

    fact[0] = 1;
    fact[1] = 1;
    for (int i = 2; i < 202020; i++)
    {
        fact[i] = (fact[i - 1] * i) % mod;
    }

    int T;
    std::cin >> T;
    while (T--)
    {
        int N;
        std::cin >> N;
        std::vector<ll> v(N);
        std::map<ll, ll> mp;
        ll AND = __LONG_LONG_MAX__;
        ll x = 1;

        for (int i = 0; i < N; i++)
        {
            std::cin >> v[i];
            AND &= v[i];
            mp[v[i]]++;
            x *= mp[v[i]];
            x %= mod;
        }

        ll res = 0;
        ll cnt = 0;

        for (int i = 0; i < N; i++)
        {
            if ((AND == v[i]))
            {
                cnt++;
            }
        }

        if (cnt >= 2)
        {
            res = ((cnt - 1) * cnt) % mod;
            res = (res * fact[N - 2]) % mod;
        }

        std::cout << res % mod << "\n";
    }

    return 0;
}
```