#include <bits/stdc++.h>
using namespace std;
struct _IO {
  _IO() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
} _io;
const long long maxn = 1e6 + 10, mod = 1e9 + 7;
template <typename... T>
long long madd(T... args) {
  long long res = 0;
  for (auto i : {args...}) {
    res = (res + ((i % mod + mod) % mod)) % mod;
  }
  return res % mod;
}
template <typename... T>
long long mmul(T... args) {
  long long res = 1;
  for (auto i : {args...}) {
    res = (res * ((i % mod + mod) % mod)) % mod;
  }
  return res % mod;
}
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
long long mpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res % mod;
}
long long _inv(long long x) { return mpow(x, mod - 2); }
void solve() {
  long long cnt0 = 0, cnt1 = 0;
  long long n;
  cin >> n;
  long long t;
  while (n--) {
    cin >> t;
    if (t == 1)
      ++cnt1;
    else if (t == 0)
      ++cnt0;
  }
  cout << cnt1 * qpow(2ll, cnt0) << '\n';
}
int main() {
  long long T = 1;
  cin >> T;
  while (T--) solve();
  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;
struct _IO {
  _IO() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
} _io;
const long long maxn = 1e6 + 10, mod = 1e9 + 7;
template <typename... T>
long long madd(T... args) {
  long long res = 0;
  for (auto i : {args...}) {
    res = (res + ((i % mod + mod) % mod)) % mod;
  }
  return res % mod;
}
template <typename... T>
long long mmul(T... args) {
  long long res = 1;
  for (auto i : {args...}) {
    res = (res * ((i % mod + mod) % mod)) % mod;
  }
  return res % mod;
}
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
long long mpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res % mod;
}
long long _inv(long long x) { return mpow(x, mod - 2); }
void solve() {
  long long cnt0 = 0, cnt1 = 0;
  long long n;
  cin >> n;
  long long t;
  while (n--) {
    cin >> t;
    if (t == 1)
      ++cnt1;
    else if (t == 0)
      ++cnt0;
  }
  cout << cnt1 * qpow(2ll, cnt0) << '\n';
}
int main() {
  long long T = 1;
  cin >> T;
  while (T--) solve();
  return 0;
}

```