#include <bits/stdc++.h>
long long q_pow(long long x, long long n, long long mod) {
  x %= mod;
  long long ret = 1;
  while (n) {
    if (n & 1) ret = x * ret % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ret;
}
long long getbit(long long now) {
  int cnt = 0;
  while (now) {
    cnt++;
    now -= (now) & (-now);
  }
  return cnt;
}
int dx[] = {1, 0};
int dy[] = {0, 1};
using namespace std;
void dbg_out() { cout << "\n"; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cout << ' ' << H;
  dbg_out(T...);
}
const long long maxn = 1e6 + 5;
const long long maxm = 250 + 10;
const long long INF = 0x3f3f3f3f;
const long long IINF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 998244353;
void add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
void mul(long long &x, long long y) {
  x -= y;
  if (x < 0) x += mod;
}
long long a[maxn];
map<long long, long long> g;
map<long long, long long> num;
void solve() {
  int n;
  cin >> n;
  long long cnt0 = 0;
  long long cnt1 = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 0) cnt0++;
    if (a[i] == 1) cnt1++;
  }
  cout << cnt1 * (1LL << cnt0) << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
long long q_pow(long long x, long long n, long long mod) {
  x %= mod;
  long long ret = 1;
  while (n) {
    if (n & 1) ret = x * ret % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ret;
}
long long getbit(long long now) {
  int cnt = 0;
  while (now) {
    cnt++;
    now -= (now) & (-now);
  }
  return cnt;
}
int dx[] = {1, 0};
int dy[] = {0, 1};
using namespace std;
void dbg_out() { cout << "\n"; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cout << ' ' << H;
  dbg_out(T...);
}
const long long maxn = 1e6 + 5;
const long long maxm = 250 + 10;
const long long INF = 0x3f3f3f3f;
const long long IINF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 998244353;
void add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
void mul(long long &x, long long y) {
  x -= y;
  if (x < 0) x += mod;
}
long long a[maxn];
map<long long, long long> g;
map<long long, long long> num;
void solve() {
  int n;
  cin >> n;
  long long cnt0 = 0;
  long long cnt1 = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 0) cnt0++;
    if (a[i] == 1) cnt1++;
  }
  cout << cnt1 * (1LL << cnt0) << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```