#include <bits/stdc++.h>
using namespace std;
long long mod = 22866613371;
long long p = 1000039, p1 = 911, mod1 = 999999001, INF = 1e9;
const long double PI = acos(-1);
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void dfs(vector<bool> &used, vector<vector<int>> &g, int v) {
  if (used[v]) return;
  used[v] = true;
  for (auto i : g[v]) {
    if (!used[i]) {
      dfs(used, g, i);
    }
  }
}
long long binpow(long long a, long long x) {
  if (a == 1 || x == 0) return 1;
  if (x == 1) return a;
  if (x & 1)
    return binpow(a, x - 1) * a % mod1;
  else {
    long long temp = binpow(a, x >> 1);
    return temp * temp % mod1;
  }
}
void solve() {
  long long n, x;
  cin >> n;
  long long cnt = 1;
  bool fl = false, fl1 = true;
  cin >> x;
  fl = x;
  if (x == 1) cnt++;
  for (long long i = 1; i < n; i++) {
    cin >> x;
    if (!fl && x == 0) {
      fl1 = false;
    } else if (!fl && x == 1) {
      fl = true;
      cnt++;
    } else if (fl && x == 1) {
      cnt += 5;
    } else {
      fl = false;
    }
  }
  if (fl1)
    cout << cnt << '\n';
  else
    cout << -1 << '\n';
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  long long T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
long long mod = 22866613371;
long long p = 1000039, p1 = 911, mod1 = 999999001, INF = 1e9;
const long double PI = acos(-1);
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void dfs(vector<bool> &used, vector<vector<int>> &g, int v) {
  if (used[v]) return;
  used[v] = true;
  for (auto i : g[v]) {
    if (!used[i]) {
      dfs(used, g, i);
    }
  }
}
long long binpow(long long a, long long x) {
  if (a == 1 || x == 0) return 1;
  if (x == 1) return a;
  if (x & 1)
    return binpow(a, x - 1) * a % mod1;
  else {
    long long temp = binpow(a, x >> 1);
    return temp * temp % mod1;
  }
}
void solve() {
  long long n, x;
  cin >> n;
  long long cnt = 1;
  bool fl = false, fl1 = true;
  cin >> x;
  fl = x;
  if (x == 1) cnt++;
  for (long long i = 1; i < n; i++) {
    cin >> x;
    if (!fl && x == 0) {
      fl1 = false;
    } else if (!fl && x == 1) {
      fl = true;
      cnt++;
    } else if (fl && x == 1) {
      cnt += 5;
    } else {
      fl = false;
    }
  }
  if (fl1)
    cout << cnt << '\n';
  else
    cout << -1 << '\n';
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  long long T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

```