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.Scanner;
  
  public class C {
      static int[] POW10 = new int[10];
  
      public static void main(String[] args) {
          POW10[0] = 1;
          for (int i = 1; i < 10; i++) {
              POW10[i] = POW10[i - 1] * 10;
          }
          Scanner scn = new Scanner(System.in);
          int t = scn.nextInt();
          while (t-- > 0) {
              int n = scn.nextInt();
              int k = scn.nextInt() + 1;
              int[] av = new int[10];
              int[] count = new int[10];
              int last = scn.nextInt();
              for (int i = 1; i < n; i++) {
                  int val = scn.nextInt();
                  av[last] = POW10[val - last] - 1;
                  last = val;
              }
              for (int i = 0; i < last; i++) {
                  if (k <= av[i]) {
                      count[i] = k;
                      k = 0;
                      break;
                  } else {
                      count[i] = av[i];
                      k -= av[i];
                  }
              }
              count[last] = k;
              long ans = 0;
              for (int i = 0; i < 10; i++) {
                  ans += 1l * POW10[i] * count[i];
              }
              System.out.println(ans);
  
          }
      }
  }
  