#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
long long fastpower(long long x, long long n, long long M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return fastpower((x * x) % M, n / 2, M);
  else
    return (x * fastpower((x * x) % M, (n - 1) / 2, M)) % M;
}
long long GCD(long long A, long long B) {
  if (B == 0)
    return A;
  else
    return GCD(B, A % B);
}
bool vowl(char c) {
  return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
long long modInverse(long long A, long long M) {
  return fastpower(A, M - 2, M);
}
void sieve(long long N) {
  bool isPrime[N + 1];
  for (long long i = 0; i <= N; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (long long i = 2; i * i <= N; ++i) {
    if (isPrime[i] == true) {
      for (long long j = i * i; j <= N; j += i) isPrime[j] = false;
    }
  }
}
vector<long long> factorize(long long n) {
  vector<long long> res;
  for (long long i = 2; i * i <= n; ++i) {
    while (n % i == 0) {
      res.push_back(i);
      n /= i;
    }
  }
  if (n != 1) {
    res.push_back(n);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long arr[n];
    for (long long i = 0; i < n; i++) cin >> arr[i];
    long long one = 0, zero = 0;
    for (long long i = 0; i < n; i++) {
      if (arr[i] == 1) one++;
      if (arr[i] == 0) zero++;
    }
    long long ans = ((long long)1 << zero);
    ans *= (long long)one;
    cout << ans << endl;
  }
  return 0;
}

### 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>
using namespace std;
const int M = 1e9 + 7;
long long fastpower(long long x, long long n, long long M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return fastpower((x * x) % M, n / 2, M);
  else
    return (x * fastpower((x * x) % M, (n - 1) / 2, M)) % M;
}
long long GCD(long long A, long long B) {
  if (B == 0)
    return A;
  else
    return GCD(B, A % B);
}
bool vowl(char c) {
  return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
long long modInverse(long long A, long long M) {
  return fastpower(A, M - 2, M);
}
void sieve(long long N) {
  bool isPrime[N + 1];
  for (long long i = 0; i <= N; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (long long i = 2; i * i <= N; ++i) {
    if (isPrime[i] == true) {
      for (long long j = i * i; j <= N; j += i) isPrime[j] = false;
    }
  }
}
vector<long long> factorize(long long n) {
  vector<long long> res;
  for (long long i = 2; i * i <= n; ++i) {
    while (n % i == 0) {
      res.push_back(i);
      n /= i;
    }
  }
  if (n != 1) {
    res.push_back(n);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long arr[n];
    for (long long i = 0; i < n; i++) cin >> arr[i];
    long long one = 0, zero = 0;
    for (long long i = 0; i < n; i++) {
      if (arr[i] == 1) one++;
      if (arr[i] == 0) zero++;
    }
    long long ans = ((long long)1 << zero);
    ans *= (long long)one;
    cout << ans << endl;
  }
  return 0;
}

```