#include <bits/stdc++.h>
using namespace std;
void solution() {
  long long n, ans = 1;
  cin >> n;
  bool chk = 0;
  vector<long long> a(n);
  for (long long i = 0; i < n; i += 1) cin >> a[i];
  for (long long i = 0; i < n; i += 1) {
    if (a[i])
      ans += ((i > 0 && a[i - 1]) ? 5 : 1);
    else if (i > 0 && !a[i - 1]) {
      chk = 1;
      break;
    }
  }
  cout << (chk ? -1 : ans) << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solution();
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

Petya has got an interesting flower. Petya is a busy person, so he sometimes forgets to water it. You are given n days from Petya's live and you have to determine what happened with his flower in the end.

The flower grows as follows: 

  * If the flower isn't watered for two days in a row, it dies. 
  * If the flower is watered in the i-th day, it grows by 1 centimeter. 
  * If the flower is watered in the i-th and in the (i-1)-th day (i > 1), then it grows by 5 centimeters instead of 1. 
  * If the flower is not watered in the i-th day, it does not grow. 



At the beginning of the 1-st day the flower is 1 centimeter tall. What is its height after n days?

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The first line of each test case contains the only integer n (1 ≤ n ≤ 100).

The second line of each test case contains n integers a_1, a_2, ..., a_n (a_i = 0 or a_i = 1). If a_i = 1, the flower is watered in the i-th day, otherwise it is not watered.

Output

For each test case print a single integer k — the flower's height after n days, or -1, if the flower dies.

Example

Input


4
3
1 0 1
3
0 1 1
4
1 0 0 1
1
0


Output


3
7
-1
1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void solution() {
  long long n, ans = 1;
  cin >> n;
  bool chk = 0;
  vector<long long> a(n);
  for (long long i = 0; i < n; i += 1) cin >> a[i];
  for (long long i = 0; i < n; i += 1) {
    if (a[i])
      ans += ((i > 0 && a[i - 1]) ? 5 : 1);
    else if (i > 0 && !a[i - 1]) {
      chk = 1;
      break;
    }
  }
  cout << (chk ? -1 : ans) << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solution();
  return 0;
}

```