description: In Berland, n different types of banknotes are used. Banknotes of the i-th type have denomination 10^{a_i} burles (burles are the currency used in Berland); the denomination of banknotes of the first type is exactly 1.
  
  Let's denote f(s) as the minimum number of banknotes required to represent exactly s burles. For example, if the denominations of banknotes used in Berland are 1, 10 and 100, then f(59) = 14: 9 banknotes with denomination of 1 burle and 5 banknotes with denomination of 10 burles can be used to represent exactly 9 ⋅ 1 + 5 ⋅ 10 = 59 burles, and there's no way to do it with fewer banknotes.
  
  For a given integer k, find the minimum positive number of burles s that cannot be represented with k or fewer banknotes (that is, f(s) > k).
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
  
  The first line of each test case contains two integers n and k (1 ≤ n ≤ 10; 1 ≤ k ≤ 10^9).
  
  The next line contains n integers a_1, a_2, ..., a_n (0 = a_1 < a_2 < ... < a_n ≤ 9).
  
  Output
  
  For each test case, print one integer — the minimum positive number of burles s that cannot be represented with k or fewer banknotes.
  
  Example
  
  Input
  
  
  4
  3 13
  0 1 2
  2 777
  0 4
  3 255
  0 1 3
  10 1000000000
  0 1 2 3 4 5 6 7 8 9
  
  
  Output
  
  
  59
  778
  148999
  999999920999999999
solution: import java.util.Scanner;
  
  public class Simple{  
      public static void main(String args[]){  
          //System.out.println("Hello Java");
          Scanner s = new Scanner(System.in);
          int t = s.nextInt();
          while (t>0){
              int n = s.nextInt();
              int k = s.nextInt();
              int arr[] = new int[n];
              long[] arr2 = new long[n];
              for(int i=0;i<n;i++)arr[i]=s.nextInt();
              for(int i=0;i<n;i++){
                  arr2[i] = (long)Math.pow(10,(arr[i]));
              }
  
  
              int max_curr = n;
              long dp[] = new long[n];
              for(int i=0;i<n-1;i++){
                  //long ul = (long)Math.pow(10, arr[i+1]);
                  //System.out.println(arr2[i]);
                  dp[i+1] = (arr2[i+1]-1)/arr2[i] + dp[i];
  
                  if(dp[i+1]>k){
                      max_curr = i;
                      break;
                  }
  
              }
              long ans =0;
              if(max_curr==n){
                  long k1 = (long)k;
                  k1 = k1 - (dp[n-1]-1);
  
                  ans = k1*(arr2[n-1]) + (arr2[n-1]-1);
              }
              else{
                  long k1 = (long)k;
                  k1 = k1 - (dp[max_curr]-1);
                  ans = k1*arr2[max_curr] + (arr2[max_curr]-1);
                  //System.out.println(k1);
              }
              System.out.println(ans);
  
              // for(int i=0;i<n;i++){
              //     System.out.print(arr2[i]+" ");
              // }
              // System.out.println();
              // for(int i=0;i<n;i++){
              //     System.out.print(dp[i]+" ");
              // }
              t--;
          }  
      }  
  }  