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: //package Div2.C;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  
  public class Banknotes {
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          int t=Integer.parseInt(br.readLine());
          while(t>0){
              String []nk=br.readLine().split(" ");
              int n=Integer.parseInt(nk[0]);
              int k=Integer.parseInt(nk[1]);
              String []str=br.readLine().split(" ");
              int []a=new int[n];
              for(int i=0;i<n;i++)
                  a[i]=Integer.parseInt(str[i]);
              long s=0;
              int bk = k+1;//bank notes to represent
              for(int i=0;i<n && bk>0;i++){
                  long dmv = (long)Math.pow(10,a[i]);
                  //if there is higher denomination available
                  if(i+1<n){
                      //max bank notes of a[i] denomination
                      int mbn = (int)Math.pow(10,a[i+1]-a[i])-1;
                      //denomination value
                      s += Integer.min(bk, mbn) * dmv;
                      bk -= Integer.min(bk, mbn);
                  }else{
                      s += bk * dmv;
                      bk = 0;
                  }
              }
              System.out.println(s);
              t--;
          }
      }
  }
  