#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
using namespace chrono;
void ip(vector<long long int> &arr) {
  for (long long int i = 0; i < arr.size(); i++) {
    cin >> arr[i];
  }
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> arr(n);
  long long int sum = 0;
  map<long long int, long long int> m;
  for (long long int i = 0; i < n; i++) {
    cin >> arr[i];
    m[arr[i]]++;
  }
  long long int z = m[0];
  long long int ans = ((1LL << z)) * (m[1]);
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  auto start1 = high_resolution_clock::now();
  long long int tests;
  cin >> tests;
  while (tests--) {
    solve();
  }
  auto stop1 = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop1 - start1);
  cerr << "Time: " << duration.count() / 1000 << "\n";
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
using namespace chrono;
void ip(vector<long long int> &arr) {
  for (long long int i = 0; i < arr.size(); i++) {
    cin >> arr[i];
  }
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> arr(n);
  long long int sum = 0;
  map<long long int, long long int> m;
  for (long long int i = 0; i < n; i++) {
    cin >> arr[i];
    m[arr[i]]++;
  }
  long long int z = m[0];
  long long int ans = ((1LL << z)) * (m[1]);
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  auto start1 = high_resolution_clock::now();
  long long int tests;
  cin >> tests;
  while (tests--) {
    solve();
  }
  auto stop1 = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop1 - start1);
  cerr << "Time: " << duration.count() / 1000 << "\n";
}

```