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.io.*;
  
  public class Solution{
      static FastScanner sc;
      public static void main (String[] args) throws IOException {
          sc = new FastScanner();
          int t = sc.ni();
          while(t-->0){
              int n = sc.ni();
              int k = sc.ni() + 1;
              int[] a = new int[n];
              for(int i = 0; i<n; i++){
                  a[i] = sc.ni();
              }
              long sum = 0;
              for(int i = 0; i<n-1; i++){
                  int quantity = (int)Math.min(k, Math.pow(10, a[i+1]) / Math.pow(10, a[i]) - 1);
                  sum += Math.pow(10, a[i]) * quantity;
                  k-=quantity;
              }
              sum += (long)k * (long)Math.pow(10, a[a.length-1]);
              System.out.println(sum);
          }
      }
      static class FastScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public FastScanner() {
              br = new BufferedReader(new InputStreamReader(System.in));
              st = null;
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int ni() {
              return Integer.parseInt(next());
          }
          long nl() { return Long.parseLong(next());}
      }
  }
  