#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int64_t> a(n);
    for (auto &x : a) cin >> x;
    int64_t sum = accumulate(a.begin(), a.end(), 0LL);
    int64_t c1 = count(a.begin(), a.end(), 1);
    int64_t c0 = count(a.begin(), a.end(), 0);
    if (sum == 0) {
      cout << "0\n";
    } else {
      int64_t ans = c1;
      for (int i = 0; i < c0; ++i) ans *= 2LL;
      cout << ans << '\n';
    }
  }
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int64_t> a(n);
    for (auto &x : a) cin >> x;
    int64_t sum = accumulate(a.begin(), a.end(), 0LL);
    int64_t c1 = count(a.begin(), a.end(), 1);
    int64_t c0 = count(a.begin(), a.end(), 0);
    if (sum == 0) {
      cout << "0\n";
    } else {
      int64_t ans = c1;
      for (int i = 0; i < c0; ++i) ans *= 2LL;
      cout << ans << '\n';
    }
  }
}

```