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.io.*;
  import java.util.*;
  
  public class q3 {
      public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  
      public static void solve() throws Exception {
          String[] parts = br.readLine().split(" ");
          int n = Integer.parseInt(parts[0]);
          int k = Integer.parseInt(parts[1]) + 1;
  
          Long[] arr = new Long[n];
  
          parts = br.readLine().split(" ");
          for (int i = 0; i < n; i++) {
              arr[i] = (long)Math.pow(10,Integer.parseInt(parts[i]));
          }
  
          long ans = 0;
          for(int i = 0;i < arr.length - 1;i++){
              long val = (arr[i + 1] / arr[i]) - 1;
  
              if(val > k){
                  ans += arr[i] * k;
                  k = 0;
                  break;
              }else{
                  ans += arr[i] * val;
                  k -= val;
              }
          }
          ans += arr[arr.length - 1] * k;
          System.out.println(ans);
  
      }
  
      public static void main(String[] args) throws Exception {
  
          int tests = Integer.parseInt(br.readLine());
          for (int test = 1; test <= tests; test++) {
              solve();
          }
      }
  }
  