#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
template <typename T, typename U>
long long modMul(T a, U b) {
  return ((a % 1000000007) * 1LL * (b % 1000000007)) % 1000000007;
}
template <typename T, typename U>
long long binpower(T n, U k) {
  T res = 1;
  while (k) {
    if (k & 1) {
      res = modMul(res, n);
    }
    n = modMul(n, n);
    k >>= 1;
  }
  return res % 1000000007;
}
template <typename T>
T moduloInverse(T a) {
  return binpower(a, 1000000007 - 2);
}
template <typename T, typename U>
long long modAdd(T a, U b) {
  return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
}
template <typename T, typename U>
long long modSub(T a, U b) {
  return ((a % 1000000007) - (b % 1000000007) + 1000000007) % 1000000007;
}
template <typename T, typename U>
long long modDiv(T a, U b) {
  return modMul(a, moduloInverse(b));
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
void _print(long long x) { cerr << x; }
void _print(float x) { cerr << x; }
void _print(double x) { cerr << x; }
void _print(string x) { cerr << x; }
void _print(unsigned long long x) { cerr << x; }
void _print(char x) { cerr << x; }
void _print(bool x) { cerr << x; }
template <typename T>
vector<T> inp(long long n) {
  vector<T> ain(n);
  for (long long i = 0; i < n; i++) cin >> ain[i];
  return ain;
}
template <typename T>
void print(vector<T> arr) {
  for (long long i = 0; i < arr.size(); i++) cout << arr[i] << " ";
  cout << "\n";
  return;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> arr = inp<long long>(n);
  vector<long long> brr;
  long long h = 1;
  long long prev = -1;
  long long ct = 0;
  for (long long i = 0; i < n; i++) {
    if (arr[i]) {
      if (i > 0) {
        if (arr[i - 1]) {
          h += 5;
        } else {
          h += 1;
        }
      } else {
        h++;
      }
    } else {
      if (i > 0) {
        if (arr[i - 1] == 0) {
          h = -1;
          break;
        }
      }
    }
  }
  cout << h << "\n";
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long 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;
using namespace std::chrono;
template <typename T, typename U>
long long modMul(T a, U b) {
  return ((a % 1000000007) * 1LL * (b % 1000000007)) % 1000000007;
}
template <typename T, typename U>
long long binpower(T n, U k) {
  T res = 1;
  while (k) {
    if (k & 1) {
      res = modMul(res, n);
    }
    n = modMul(n, n);
    k >>= 1;
  }
  return res % 1000000007;
}
template <typename T>
T moduloInverse(T a) {
  return binpower(a, 1000000007 - 2);
}
template <typename T, typename U>
long long modAdd(T a, U b) {
  return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
}
template <typename T, typename U>
long long modSub(T a, U b) {
  return ((a % 1000000007) - (b % 1000000007) + 1000000007) % 1000000007;
}
template <typename T, typename U>
long long modDiv(T a, U b) {
  return modMul(a, moduloInverse(b));
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
void _print(long long x) { cerr << x; }
void _print(float x) { cerr << x; }
void _print(double x) { cerr << x; }
void _print(string x) { cerr << x; }
void _print(unsigned long long x) { cerr << x; }
void _print(char x) { cerr << x; }
void _print(bool x) { cerr << x; }
template <typename T>
vector<T> inp(long long n) {
  vector<T> ain(n);
  for (long long i = 0; i < n; i++) cin >> ain[i];
  return ain;
}
template <typename T>
void print(vector<T> arr) {
  for (long long i = 0; i < arr.size(); i++) cout << arr[i] << " ";
  cout << "\n";
  return;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> arr = inp<long long>(n);
  vector<long long> brr;
  long long h = 1;
  long long prev = -1;
  long long ct = 0;
  for (long long i = 0; i < n; i++) {
    if (arr[i]) {
      if (i > 0) {
        if (arr[i - 1]) {
          h += 5;
        } else {
          h += 1;
        }
      } else {
        h++;
      }
    } else {
      if (i > 0) {
        if (arr[i - 1] == 0) {
          h = -1;
          break;
        }
      }
    }
  }
  cout << h << "\n";
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```