#include <bits/stdc++.h>
using namespace std;
int main() {
  int T, i, j, k = 0;
  cin >> T;
  while (T-- > 0) {
    int n;
    cin >> n;
    int a[n + 1], d = 1, die = 0;
    for (i = 1; i < n + 1; i++) cin >> a[i];
    a[0] = 0;
    for (i = 1; i < n + 1; i++) {
      if (a[i] == 1) {
        d++;
        if (a[i - 1] == 1) d += 4;
      } else {
        if (a[i - 1] == 0 && i > 1) {
          die++;
          break;
        }
      }
    }
    if (die != 0)
      cout << -1 << endl;
    else
      cout << d << endl;
  }
}

### Prompt

Please formulate 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;
int main() {
  int T, i, j, k = 0;
  cin >> T;
  while (T-- > 0) {
    int n;
    cin >> n;
    int a[n + 1], d = 1, die = 0;
    for (i = 1; i < n + 1; i++) cin >> a[i];
    a[0] = 0;
    for (i = 1; i < n + 1; i++) {
      if (a[i] == 1) {
        d++;
        if (a[i - 1] == 1) d += 4;
      } else {
        if (a[i - 1] == 0 && i > 1) {
          die++;
          break;
        }
      }
    }
    if (die != 0)
      cout << -1 << endl;
    else
      cout << d << endl;
  }
}

```