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.lang.*;
  import java.util.*;
  import java.io.*;
  import java.math.*;
  
  public class Main {
  
      static void deal(int n,int k,int[] arr) {
          long res = 0;
          k++;
          for(int i=0;i<n-1;i++) {
              int t = (int)Math.pow(10,arr[i+1]-arr[i])-1;
  //          System.out.println(String.format("%d %d %d",t,k,res));
              if(t<k) {
                  k -= t;
                  res += (long)Math.pow(10,arr[i])*t;
              } else {
                  res += (long)Math.pow(10,arr[i])*k;
                  out.println(res);
                  return;
              }
          }
          res += (long)Math.pow(10,arr[n-1])*k;
          out.println(res);
      }
      
  	public static void main(String[] args) {
          MyScanner sc = new MyScanner();
          out = new PrintWriter(new BufferedOutputStream(System.out));
          int t = sc.nextInt();
          for(int i=0;i<t;i++) {
              int n = sc.nextInt();
              int k = sc.nextInt();
              int[] arr = new int[n];
              for(int j=0;j<n;j++) arr[j] = sc.nextInt();
              deal(n,k,arr);
          }
          out.close();
      }
      
      //-----------PrintWriter for faster output---------------------------------
      public static PrintWriter out;
      
      //-----------MyScanner class for faster input----------
      public static class MyScanner {
          BufferedReader br;
          StringTokenizer st;
          
          public MyScanner() {
                  br = new BufferedReader(new InputStreamReader(System.in));
          }
          
          String next() {
                  while (st == null || !st.hasMoreElements()) {
                          try {
                                  st = new StringTokenizer(br.readLine());
                          } catch (IOException e) {
                                  e.printStackTrace();
                          }
                  }
                  return st.nextToken();
          }
          
          int nextInt() {
                  return Integer.parseInt(next());
          }
          
          long nextLong() {
                  return Long.parseLong(next());
          }
          
          double nextDouble() {
                  return Double.parseDouble(next());
          }
          
          String nextLine(){
                  String str = "";
      try {
              str = br.readLine();
      } catch (IOException e) {
              e.printStackTrace();
      }
      return str;
          }
          
      }
  }