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;
  long long N = 2e5;
  pair<long long, string> pa[100000];
  vector<long long> v, vv, v2;
  map<long long, long long> mp;
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      v.clear();
      long long n, cnt = 1;
      bool q = 0;
      cin >> n;
      for (long long i = 0; i < n; i++) {
        long long x;
        cin >> x;
        v.push_back(x);
      }
      if (n == 1 && v[0] == 1)
        cout << 2 << endl;
      else if (n == 1 && v[0] == 0)
        cout << 1 << endl;
      else {
        if (v[0] == 1) cnt++;
        for (long long i = 1; i < n; i++) {
          if ((v[i] == 1 && v[i - 1] == 0)) {
            cnt++;
          } else if (v[i] == 0 && v[i - 1] == 0) {
            q = 1;
            break;
          } else if (v[i] == 1 && v[i - 1] == 1) {
            cnt += 5;
          }
        }
        if (q)
          cout << -1 << endl;
        else
          cout << cnt << endl;
      }
    }
  }
  