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 long long MAXN = 1e7 + 10;
  long long factory[] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
  long long visited[MAXN] = {0};
  int a[MAXN];
  long long quick_m(long long a, long long b, long long p) {
    long long res = 1 % p;
    while (b) {
      if (b & 1) res = res * a % p;
      b >>= 1;
      a = a * a % p;
    }
    return res;
  }
  bool Cantor(int str[], int n) {
    long long result = 0;
    for (int i = 0; i < n; i++) {
      int counted = 0;
      for (int j = i + 1; j < n; j++) {
        if (str[i] > str[j]) {
          counted++;
        }
      }
      result += counted * factory[n - i - 1];
    }
    if (!visited[result]) {
      visited[result] = 1;
      return 1;
    }
    return 0;
  }
  int main() {
    int n;
    cin >> n;
    while (n--) {
      int m;
      cin >> m;
      for (int i = 0; i < m; i++) {
        cin >> a[i];
      }
      int sum = 1;
      int flag = 0;
      int ans = 0;
      for (int i = 0; i < m; i++) {
        if (a[i] == 1) {
          ans = 0;
          if (flag == 1) {
            sum += 5;
          } else if (flag == 0) {
            sum += 1;
            flag = 1;
          }
        } else if (a[i] == 0) {
          flag = 0;
          ans++;
        }
        if (ans == 2) {
          sum = -1;
          break;
        }
      }
      cout << sum << endl;
    }
    return 0;
  }
  