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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  public class Main {
  
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              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;
          }
      }
  
  
      public static void main(String[] args) {
  
          FastReader sc = new FastReader();
  
          int t = sc.nextInt();
          for (int i = 0; i < t; i++) {
              int n = sc.nextInt();
              long k = sc.nextInt();
              int[] vals = new int[n];
              long[] anss = new long[n];
              for (int j = 0 ; j < n; j++) {
                  vals[j] = sc.nextInt();
              }
  
              for (int j = 0 ; j < n ; j++) {
  
                  if ( j == n - 1) {
                      anss[j] = k +1;
                      break ;
                  }
                  long temp = power(10,vals[j+1] - vals[j]) - 1;
                  if (temp > k) {
                      anss[j] = k + 1 ;
                      break ;
                  }
                  k = k - temp ;
                  anss[j] = temp ;
              }
  
              long ans  = 0;
              for (int j = 0 ; j < n ; j++) {
                  ans = ans + power(10,vals[j]) * anss[j] ;
              }
  
              System.out.println(ans);
  
          }
  
      }
  
      static long power(long x, long y)
      {
          long res = 1;
          x = x ;
          if (x == 0)
              return 0;
  
          while (y > 0)
          {
              if ((y & 1) != 0)
                  res = (res * x) ;
  
              y = y >> 1; // y = y/2
              x = (x * x) ;
          }
          return res;
      }
  }