human
PYTHON3 solution for "Luntik came out for a morning stroll and found an array a of length n. He calculated the sum s of the elements of the array (s= ∑_{i=1}^{n} a_i). Luntik calls a subsequence of the array a nearly full if the sum of the numbers in that subsequence is equal to s-1.

Luntik really wants to know the number of nearly full subsequences of the array a. But he needs to come home so he asks you to solve that problem!

A sequence x is a subsequence of a sequence y if x can be obtained from y by deletion of several (possibly, zero or all) elements.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The next 2 ⋅ t lines contain descriptions of test cases. The description of each test case consists of two lines.

The first line of each test case contains a single integer n (1 ≤ n ≤ 60) — the length of the array.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9) — the elements of the array a.

Output

For each test case print the number of nearly full subsequences of the array.

Example

Input

5
5
1 2 3 4 5
2
1000 1000
2
1 0
5
3 0 2 1 1
5
2 1 0 3 0

Output

1
0
2
4
4

Note

In the first test case, s=1+2+3+4+5=15, only (2,3,4,5) is a nearly full subsequence among all subsequences, the sum in it is equal to 2+3+4+5=14=15-1.

In the second test case, there are no nearly full subsequences.

In the third test case, s=1+0=1, the nearly full subsequences are (0) and () (the sum of an empty subsequence is 0).":



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Write me a Python program to print the number of nearly full subsequences of the array?
assistant
#Here is Python program to print the number of nearly full subsequences of the array.
 def nCr(n,r):
    return fact(n) / (fact(r) * fact(n - r))
def fact(n):
    res = 1
    for i in range(2,n+1):
        res*=i
    return res
def solve():
    n = int(input())
    arr = list()
    cnt1,cnt0 = 0,0
    arr = L()
    for i in range(n):
        if arr[i] == 1:
            cnt1+=1
        elif arr[i] == 0:
            cnt0+=1
    if cnt0>0 and cnt1:
        print(pow(2,cnt0)*cnt1)
    else:
        print(cnt1)
S=lambda:input().rstrip();
I=lambda:int(S());
M=lambda:map(int,S().split());
L=lambda:list(M());
H=1000000000+7
for _ in range(I()):
    solve()