#include <bits/stdc++.h>
using namespace std;
template <typename T, typename T1>
T abs(T a, T1 b) {
  if (b > a) return b - a;
  return a - b;
}
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
long long mul_mod(long long a, long long b, long long p) {
  return ((a % p) * (b % p)) % p;
}
long long add_mod(long long a, long long b, long long p) {
  return ((a % p) + (b % p)) % p;
}
long long sub_mod(long long a, long long b, long long p) {
  return ((a - b) % p) + ((a >= b) ? 0 : p);
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long cnt0 = 0, cnt1 = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] == 0) cnt0++;
    if (a[i] == 1) cnt1++;
  }
  cout << (long long)pow(2, cnt0) * cnt1 << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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;
template <typename T, typename T1>
T abs(T a, T1 b) {
  if (b > a) return b - a;
  return a - b;
}
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
long long mul_mod(long long a, long long b, long long p) {
  return ((a % p) * (b % p)) % p;
}
long long add_mod(long long a, long long b, long long p) {
  return ((a % p) + (b % p)) % p;
}
long long sub_mod(long long a, long long b, long long p) {
  return ((a - b) % p) + ((a >= b) ? 0 : p);
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long cnt0 = 0, cnt1 = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] == 0) cnt0++;
    if (a[i] == 1) cnt1++;
  }
  cout << (long long)pow(2, cnt0) * cnt1 << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```