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.*;
  import java.lang.*;
  import java.io.*;
  public class Main
  {
      public static void main (String[] args) throws java.lang.Exception
      {
              Scanner sc = new Scanner(System.in);
              int t = sc.nextInt();
              while(t-->0) {
                  long n = sc.nextLong();
                  long k = sc.nextLong();
                  k+=1;
                  long res = 0;
                  long arr[] = new long[(int)n];
                  for(int i=0;i<n;i++)
                      arr[i] = sc.nextLong();
                  for(int i=0;i<n-1;i++)
                  {
                      if(k>0)
                      {
                          long x = (int)Math.pow(10,arr[i+1]-arr[i])-1;
                          long value = Math.min(x,k);
                          res = res + (int)Math.pow(10,arr[i])*value;
                          k = k-value;
                      }
                  }
                  res += k*(int)Math.pow(10,arr[(int)n-1]);
                  System.out.println(res);
              }
          }
      }