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 C {
      public static void main(String[] args) throws IOException {
          FastScanner fs=new FastScanner();
          PrintWriter out = new PrintWriter(System.out);
          int T = fs.nextInt();
          for (int tt=0; tt<T; tt++) {
              
              int n = fs.nextInt();
              int k = fs.nextInt();
              int[] arr = fs.readArray(n);
  
              ruffleSort(arr);
  
              long sum = 0;
              int notes = 0;
              int ii=0;
              while (notes<=k && ii<n-1) {
                  if (notes+(long)Math.pow(10,(arr[ii+1]-arr[ii]))-1>k) {
                      sum+=(k+1-notes)*(long)Math.pow(10,arr[ii]);
                      notes=k+1;
                      break;
                  }
                  notes+=(long)Math.pow(10,(arr[ii+1]-arr[ii]))-1;
                  sum+=((long)(Math.pow(10,arr[ii+1]-arr[ii])-1)*Math.pow(10,arr[ii]));
                  ii++;
                  // System.out.println(notes+","+ sum+","+ ii);
              }
              if (notes<=k) {
                  sum+=(k+1-notes)*(long)Math.pow(10,arr[ii]);
              }
              out.println(sum);
          }
          out.close();
  
      }
  
      static final Random random=new Random();
  
      static void ruffleSort(int[] a) {
          int n=a.length;//shuffle, then sort
          for (int i=0; i<n; i++) {
              int oi=random.nextInt(n), temp=a[oi];
              a[oi]=a[i]; a[i]=temp;
          }
          Arrays.sort(a);
      }
  
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  
  }
  