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.sql.SQLOutput;
  import java.util.*;
  
  public class CF1 {
      public static void main(String[] args) {
          FastScanner sc=new FastScanner();
          int T=sc.nextInt();
          for (int tt=0; tt<T; tt++){
              long ans=0;
              int n = sc.nextInt();
              int k = sc.nextInt();
              k++;
              int arr[]= sc.readArray(n);
              int mul[]= new int [n];
              int x=1;
              int y=0;
              for (int i=0; i<n; i++){
                  while (y<arr[i]){
                      x=x*10;
                      y++;
                  }
                  mul[i]= x;
              }
              int diff []= new int [n];
  
              for(int i=n-2; i>=0; i--){
                  diff[i]= (mul[i+1]/mul[i])-1;
              }
              for (int i=0; i<n; i++){
                  if (k==0) break;
                  if (diff[i]==0) ans+= (long)((long)mul[i] * (long)k);
                  else {
                      if (diff[i]>=k){
                          ans+=(long)((long)mul[i] * (long)k);
                          break;
                      }
                      else {
                          ans+=(long)mul[i] * (long)diff[i];
                          k-= diff[i];
                      }
                  }
              }
  //            for (int i=0; i<n; i++){
  //                System.out.print(mul[i]+" ");
  //            }
  //            System.out.println();
              System.out.println(ans);
          }
      }
      static void sort(int[] a) {
          ArrayList<Integer> l=new ArrayList<>();
          for (int i:a) l.add(i);
          Collections.sort(l);
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static void sortLong(long[] a) {
          ArrayList<Long> l=new ArrayList<>();
          for (long i:a) l.add(i);
          Collections.sort(l);
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static int gcd (int n, int m){
          if (m==0) return n;
          else return gcd(m, n%m);
      }
      static class Pair{
          int x,y;
          public Pair(int x, int y){
              this.x = x;
              this.y = y;
          }
      }
      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());
          }
      }
  }