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: 
  
  import java.util.Scanner;
  
  public class A {
      public static void main(String[] args) {
          Scanner s = new Scanner(System.in);
          int lines = s.nextInt();
  
          s.nextLine();
  
          for (int i = 0; i < lines; i += 1) {
              System.out.println(solve(s.nextInt(), s));
          }
      }
  
  
      public static int solve(int n, Scanner s) {
          boolean last = s.nextInt() == 1;
          boolean t;
          int erg = last ? 2 : 1;
  
          for (int i = 1; i < n; i++) {
              t = s.nextInt() == 1;
              if (!t && !last) erg = Integer.MIN_VALUE;
              if (t) erg++;
              if (t && last) erg += 4;
              last = t;
          }
  
          return Math.max(-1, erg);
      }
  }
  