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.math.BigInteger;
  import java.util.*;
  
  public class P1606C {
    public static void main(String[] args) {
      SpeedScanner in = new SpeedScanner();
      int t = in.nextInt();
      while(t-- != 0) {
        int n = in.nextInt(), k = in.nextInt(), a[] = new int[n];
        for(int i = 0; i < n; ++i)
          a[i] = (int) Math.pow(10, in.nextInt());
  
        BigInteger ans = BigInteger.ZERO;
        int i;
        for(i = 0; i < n - 1; ++i) {
          if((a[i + 1] - a[i]) / a[i] <= k) {
            ans = ans.add(BigInteger.valueOf(a[i + 1] - a[i]));
            k -= (a[i + 1] - a[i]) / a[i];
          }
          else
            break;
        }
        ans = ans.add(BigInteger.valueOf(k + 1).multiply(BigInteger.valueOf(a[i])));
        
        System.out.println(ans.toString());
      }
    }
  
    static class SpeedScanner {
      BufferedReader in;
      StringTokenizer splitter;
    
      public SpeedScanner() {
        in = new BufferedReader(new InputStreamReader(System.in));
      }
    
      String next() {
        while(splitter == null || !splitter.hasMoreElements())
          try {
            splitter = new StringTokenizer(in.readLine());
          } catch (IOException e) {
            e.printStackTrace();
          }
        return splitter.nextToken();
      }
    
      String nextLine() {
        String s = "";
        try {
          s = in.readLine();
        } catch (IOException e) {
          e.printStackTrace();
        }
        return s;
      }
    
      int nextInt() {
        return Integer.parseInt(next());
      }
    
      long nextLong() {
        return Long.parseLong(next());
      }
    
      double nextDouble() {
        return Double.parseDouble(next());
      }
    }
  }
  