1513_B. AND Sequences
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. 
```python
#Question B: AND Sequences
import math #for .perm()
import sys

def answer(n, a):
    mn = min(a)
    cmn = a.count(mn)
    if cmn < 2:
        #print('cmn<2')
        return 0
    # at least 2 of the smallest number
    #mnperm = math.perm(cmn, 2)
    mnperm = cmn * (cmn-1)
    mdperm = 1
    if n-2 > 0:
        #mdperm = math.perm(n-2, n-2)
        #mdperm = math.factorial(n-2) #slow
        for i in range(2, n-1):
            mdperm = mdperm * i % (10**9 +7) # keep f smaller than otherwise.
    
    #print('mnperm=', mnperm, 'mdperm=', mdperm)
    if mn == 0:
        #all other numbers irrelevant
        #print('mn=0')
        return (mnperm * mdperm) % (10**9 + 7)
    # mn > 0
    # Check if min val is binary rightward subset of all numbers. If so, nperm is the answer. That or 0.
    for i in range(n):
        if a[i] & mn != mn:
            #print('a[i] & mn != mn')
            return 0
    #print('mnperm=', mnperm, 'mdperm=', mdperm)
    return (mnperm * mdperm) % (10**9 + 7)

def main():
    t = int(input())
    while t:
        n = int(sys.stdin.readline())
        a = [int(i) for i in sys.stdin.readline().split()]
        print(answer(n, a))
        t -= 1
    return
main()
```