description: 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
solution: #include <bits/stdc++.h>
  using namespace std;
  const int mod = 1e9 + 7;
  int main() {
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      long long A[n];
      long long ans = 0;
      for (int i = 0; i <= n - 1; i++) {
        cin >> A[i];
        if (i > 0 && A[i] == 0 && A[i - 1] == 0) {
          ans = -1;
        }
      }
      if (ans == -1) {
        cout << ans << endl;
      } else {
        ans = 1;
        long long s = 0;
        for (int i = 0; i <= n - 1; i++) {
          if (A[i] == 1) {
            s++;
            if (s == 1) {
              ans++;
            } else {
              ans += 5;
            }
          } else {
            s = 0;
          }
        }
        cout << ans << endl;
      }
    }
    return 0;
  }
  