#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
const long long int maxn = 2e5 + 16, md = 1e9 + 7, inf = 2e16;
long long int gcd(long long int a, long long int b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  return gcd(b, a % b);
}
inline long long int tav(long long int n, long long int k) {
  long long int res = 1;
  while (k > 0) {
    if (k & 1) {
      res *= n;
      res %= md;
    }
    n *= n;
    n %= md;
    k >>= 1;
  }
  return res;
}
long long int a[maxn];
void solve() {
  long long int n, cur = 1;
  bool dead = false;
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    if (i) {
      if (!a[i] && !a[i - 1]) {
        dead = true;
      }
    }
    cur += a[i];
    cur += 4 * (a[i] == 1 && a[i - 1] == 1);
  }
  if (dead) {
    cout << "-1\n";
    return;
  }
  cout << cur << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int T;
  cin >> 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;
#pragma GCC optimize("Ofast")
const long long int maxn = 2e5 + 16, md = 1e9 + 7, inf = 2e16;
long long int gcd(long long int a, long long int b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  return gcd(b, a % b);
}
inline long long int tav(long long int n, long long int k) {
  long long int res = 1;
  while (k > 0) {
    if (k & 1) {
      res *= n;
      res %= md;
    }
    n *= n;
    n %= md;
    k >>= 1;
  }
  return res;
}
long long int a[maxn];
void solve() {
  long long int n, cur = 1;
  bool dead = false;
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    if (i) {
      if (!a[i] && !a[i - 1]) {
        dead = true;
      }
    }
    cur += a[i];
    cur += 4 * (a[i] == 1 && a[i - 1] == 1);
  }
  if (dead) {
    cout << "-1\n";
    return;
  }
  cout << cur << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

```