#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 10;
long long dp[N], a[] = {1, 2, 5};
void solve() {
  long long n;
  scanf("%lld", &n);
  long long ans = 1, f = 0, y = 3;
  for (long long i = 0; i <= n - 1; i++) {
    long long x;
    scanf("%lld", &x);
    if (x == 1 && y == 1)
      ans += 5;
    else if (x == 1)
      ans++;
    else if (x == 0 && y == 0)
      f = 1;
    y = x;
  }
  if (f)
    puts("-1");
  else
    cout << ans << '\n';
}
signed main() {
  long long t = 1;
  scanf("%lld", &t);
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
const long long N = 1e5 + 10;
long long dp[N], a[] = {1, 2, 5};
void solve() {
  long long n;
  scanf("%lld", &n);
  long long ans = 1, f = 0, y = 3;
  for (long long i = 0; i <= n - 1; i++) {
    long long x;
    scanf("%lld", &x);
    if (x == 1 && y == 1)
      ans += 5;
    else if (x == 1)
      ans++;
    else if (x == 0 && y == 0)
      f = 1;
    y = x;
  }
  if (f)
    puts("-1");
  else
    cout << ans << '\n';
}
signed main() {
  long long t = 1;
  scanf("%lld", &t);
  while (t--) {
    solve();
  }
  return 0;
}

```