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.math.BigInteger;
  import java.util.ArrayList;
  import java.util.List;
  import java.util.StringTokenizer;
  
  public class C {
      static class RealScanner {
          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());
          }
      }
  
      public static void main(String[] args) {
          RealScanner sc = new RealScanner();
          int t = sc.nextInt();
          while (t-- > 0) {
              long n, x;
              n = sc.nextLong();
              x = sc.nextLong() + 1;
              long res = 0;
              List<Long> l = new ArrayList<>();
              for (long i = 0; i < n; i++) {
                  l.add(sc.nextLong());
              }
              BigInteger m = new BigInteger(String.valueOf(10));
              for (long i = 0; i < n - 1; i++) {
                  if (x > 0) {
                      long val1 = (long) (Math.pow(10, l.get((int) (i + 1)) - l.get((int) i)) - 1);
                      long val2 = Math.min(x, val1);
                      res += val2 * Math.pow(10, l.get((int) i));
                      x = x - val2;
                  }
              }
  
              //long to BigInteger.. this is awesome by Java
  
              BigInteger x1 = BigInteger.valueOf(x);
              BigInteger b = BigInteger.valueOf(res);
              BigInteger pow = BigInteger.valueOf(10);
              //res += (x * Math.pow(10, l.get((int) (n - 1))));
              b = b.add(x1.multiply(pow.pow(Math.toIntExact(l.get((int) (n - 1))))));
              System.out.println(b);
          }
      }
  }
  