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;
  void Start_Crushing() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
  }
  void solve() {
    int n;
    cin >> n;
    bool dies = false;
    int ans = 1;
    for (int i = 0, entry, prev; i < n and cin >> entry; i++) {
      ans += entry;
      if (i) {
        if (not entry and not prev)
          dies = true;
        else if (entry and prev)
          ans += 4;
      }
      prev = entry;
    }
    cout << (dies ? -1 : ans);
  }
  int main() {
    Start_Crushing();
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
      if (!t) break;
      cout << "\n";
    }
    return 0;
  }
  