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.*;
  public class Life_of_a_Flower {
  
  	public static void main(String[] args) {
  		Scanner Scan = new Scanner(System.in);
  		
  
  		int t = Scan.nextInt();
  		
  		for(int i=0; i<t; i++) {
  			int flower[]= new int[10000];
  			int k=1;
  			int f=1;
  
  			int n= Scan.nextInt();
  			for(int o=0; o<n; o++) {
  
  			flower[o]= Scan.nextInt();
  			}
  			
  			for(int j=0; j<n; j++) {
  				if(j>0) {
  				    if(flower[j]==0 && flower[j-1]==0) {
  						k=-1;
  						break;}
  				    if(flower[j]==1 && flower[j-1]==1)
  					    k+=5;
  				    else if(flower[j]==1 && flower[j-1]!=1)
  					    k++;
  				}
  				else
  				{
  					if(flower[j]==1)
  						k++;
  				}
  			}
  			System.out.println(k);
  		}
  			
  	}
  
  }