human
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. 
The above is tricky. Write me a correct solution in PYTHON3.

Can you provide a Python program to find the number of permutations p of numbers ranging from 1 to n, for which the sequence a_{p_1}, a_{p_2},?
assistant
#Here is Python program to find the number of permutations p of numbers ranging from 1 to n, for which the sequence a_{p_1}, a_{p_2}, .

MOD=10**9+7
MAX=200005
fact=[1]
for zz in range(1,MAX+1):
    fact.append(((fact[-1]%MOD)*zz%MOD)%MOD) #with MOD
#    fact.append(fact[-1]*zz) #without MOD

def checkP(p):
    n=len(p)
    for i in range(n-1):
        left=2**30-1
        for j in range(i+1):
            left&=p[j]
        right=2**30-1
        for j in range(i+1,n):
        #     print(right)
            right&=p[j]
        if left!=right:
            return False
    return True

def bruteforce(a):
    a=readIntArr()
    n=len(a)
    from itertools import permutations
    ans=[]
    for p in permutations(a):
        if checkP(p):
            ans.append(p)
    return len(ans)

def main():
    
    t=int(input())
    allans=[]
    for _ in range(t):
        n=int(input())
        a=readIntArr()
        
        cumu=~0
        for x in a:
            cumu&=x
        # print('cumu:{}'.format(cumu))
        cnt=0
        for x in a:
            if x==cumu:
                cnt+=1
        # print('cnt:{}'.format(cnt))
        if cnt<=1:
            ans=0
        else:
            ans=cnt*(cnt-1)
            ans%=MOD
            ans*=fact[n-2]
            ans%=MOD
        allans.append(ans)
    multiLineArrayPrint(allans)
    
    return
    
import sys
# input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
 
def oneLineArrayPrint(arr):
    print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
    print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
    print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
 
def readIntArr():
    return [int(x) for x in input().split()]
# def readFloatArr():
#     return [float(x) for x in input().split()]

def makeArr(defaultVal,dimensionArr): # eg. makeArr(0,[n,m])
    dv=defaultVal;da=dimensionArr
    if len(da)==1:return [dv for _ in range(da[0])]
    else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
 
def queryInteractive(x,y):
    print('?...