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.*;
  import static java.lang.Math.PI;
  import static java.lang.System.in;
  import static java.lang.System.out;
  import static java.lang.System.err;
  
  public class A {    
  
      public static void main(String[] args) throws Exception {   
          Foster sc = new Foster();
          PrintWriter p = new PrintWriter(out);
  
         /*
          * Is that extra condition needed
          * Check overflow in pow function or in general
          * Check indices of read array functions
          * Think of an easier solution because the problems you solve are always easy
          * Check the iterator of loop
          */   
  
          int t = sc.nextInt();
          while(t--!=0) {
              int n = sc.nextInt();
              long k = sc.nextLong();
              long a[] = new long[n];
              for(int i = 0; i < n; i++) {
                  long x = sc.nextLong();
                  a[i] = (long)Math.pow(10, x);
              }
              long maxFree[] = new long[n];
              for(int i = 0; i < n-1; i++) {
                  maxFree[i] = a[i+1]-a[i];
              }
              long prefix[] = Arrays.copyOf(maxFree, n);
              for(int i = 1; i < n; i++) {
                  prefix[i] += prefix[i-1];
              }
              
              long prev = 0;
              k++;
              long ans = 0;
              ans = k-prev;
              prev += maxFree[0]/a[0];
              for(int i = 1; i < n; i++) {
                  if(ans >= a[i]) {
                      ans = prefix[i-1] + ((k-prev)*a[i]);
                      prev += maxFree[i]/a[i];
                  }
              }
              p.println(ans);
          }
          p.close();
      }
      
      static long[] sort(long a[]) {
          ArrayList<Long> arr = new ArrayList<>();
          for(long i : a) {
              arr.add(i);
          }
          Collections.sort(arr);
          Collections.reverse(arr);
          for(int i = 0; i < arr.size(); i++) {
              a[i] = arr.get(i);
          }
          return a;
      }
      static int[] sort(int a[]) {
          ArrayList<Integer> arr = new ArrayList<>();
          for(int i : a){
              arr.add(i);
          }
          Collections.sort(arr);
          // Collections.reverse(arr);
          for(int i = 0; i < arr.size(); i++){
              a[i] = arr.get(i);
          }
          return a;
      }   
  
      static class Foster {
          BufferedReader br = new BufferedReader(new InputStreamReader(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());
          }
          long nextLong() {
              return Long.parseLong(next());
          }
          double nextDouble() {
              return Double.parseDouble(next());
          }
          int[] intArray(int n) {
              int arr[] = new int[n];
              for(int i = 0; i < n; i++) {
                  arr[i] = nextInt();
              }
              return arr;
          }
          long[] longArray(int n) {
              long arr[] = new long[n];
              for(int i = 0; i < n; i++) {
                  arr[i] = nextLong();
              }
              return arr;
          }
          static long pow(long... a) {
              long mod = Long.MAX_VALUE;
              if(a.length == 3)   mod = a[2];
              long res = 1;
              while(a[1] > 0) {
                  if((a[1] & 1) == 1)
                      res = (res * a[0]) % mod;
                  a[1] /= 2;
                  a[0] = (a[0] * a[0]) % mod;
              }
              return res;
          }
          static long gcd(long a, long b) {
              if (b == 0)
                  return a;
              return gcd(b, a % b);
          }
      }
  }