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. 
Using python3 can you solve the prior task?

Write me 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}, .

# Author : nitish420 --------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
mod=10**9+7
# sys.setrecursionlimit(10**6)
mxm=sys.maxsize
# from functools import lru_cache
from collections import defaultdict,Counter

def factorial(n):
    dp=[1]*(n+1)
    for i in range(2,n):
        dp[i]=(dp[i-1]*i)%mod
    
    return dp

def main():
    fact=factorial(2*10**5)
    for _ in range(int(input())):
        n=int(input())
        l=list(map(int,input().split()))
        countmm=0
        mm=mxm
        for item in l:
            if item<mm:
                mm=item
                countmm=1
            elif item==mm:
                countmm+=1
        if countmm<2:
            print(0)
            continue
        nd=mm
        flag=0
        for item in l:
            if item & nd==nd:
                pass
            else:
                flag=1
                break
        if flag:
            print(0)
        else:
            print(((countmm*(countmm-1))*fact[n-2])%mod)


        

        
        
# 1
# 7
# 31 1 7 11 21 19 1

# 1
# 15
# 19 27 23 26 19 18 31 18 19 18 18 18 18 22 30



# 4
# 3
# 1 1 1
# 5
# 1 2 3 4 5
# 5
# 0 2 0 3 0
# 4
# 1 3 5 1



# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = 'x' in file.mode or 'r' not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines =...