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 Main {
      static int MOD = 998244353;
  
      // After writing solution, quick scan for:
      //   array out of bounds
      //   special cases e.g. n=1?
      //   npe, particularly in maps
      //
      // Big numbers arithmetic bugs:
      //   int overflow
      //   sorting, or taking max, after MOD
      void solve() throws IOException {
          int T = ri();
          long[] pow10 = new long[19];
          pow10[0] = 1;
          for (int i = 1; i <= 18; i++) pow10[i] = pow10[i-1] * 10;
          for (int Ti = 0; Ti < T; Ti++) {
              int[] nk = ril(2);
              int n = nk[0];
              int k = nk[1];
              int[] a = ril(n);  // contains which digits we can use
              boolean[] have = new boolean[10];
              for (int ai : a) have[ai] = true;
  
              k++;
  
              long ans = 0;
              int s = 0;
              int o = 1;
              while (k > 0 && o < 10) {
                  while (o < 10 && !have[o]) o++;
                  if (o >= 10) break;
                  int spaces = o - s;
                  long fillable = pow10[spaces] - 1;
                  long take = Math.min(k, fillable);
                  k -= take;
                  ans += take * pow10[s];
  
                  s = o;
                  o = s+1;
              }
              if (k > 0) ans += k * pow10[s];
              pw.println(ans);
          }
      }
      // IMPORTANT
      // DID YOU CHECK THE COMMON MISTAKES ABOVE?
  
      // Template code below
  
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter pw = new PrintWriter(System.out);
  
      public static void main(String[] args) throws IOException {
          Main m = new Main();
          m.solve();
          m.close();
      }
  
      void close() throws IOException {
          pw.flush();
          pw.close();
          br.close();
      }
  
      int ri() throws IOException {
          return Integer.parseInt(br.readLine().trim());
      }
  
      long rl() throws IOException {
          return Long.parseLong(br.readLine().trim());
      }
  
      int[] ril(int n) throws IOException {
          int[] nums = new int[n];
          int c = 0;
          for (int i = 0; i < n; i++) {
              int sign = 1;
              c = br.read();
              int x = 0;
              if (c == '-') {
                  sign = -1;
                  c = br.read();
              }
              while (c >= '0' && c <= '9') {
                  x = x * 10 + c - '0';
                  c = br.read();
              }
              nums[i] = x * sign;
          }
          while (c != '\n' && c != -1) c = br.read();
          return nums;
      }
  
      long[] rll(int n) throws IOException {
          long[] nums = new long[n];
          int c = 0;
          for (int i = 0; i < n; i++) {
              int sign = 1;
              c = br.read();
              long x = 0;
              if (c == '-') {
                  sign = -1;
                  c = br.read();
              }
              while (c >= '0' && c <= '9') {
                  x = x * 10 + c - '0';
                  c = br.read();
              }
              nums[i] = x * sign;
          }
          while (c != '\n' && c != -1) c = br.read();
          return nums;
      }
  
      int[] rkil() throws IOException {
          int sign = 1;
          int c = br.read();
          int x = 0;
          if (c == '-') {
              sign = -1;
              c = br.read();
          }
          while (c >= '0' && c <= '9') {
              x = x * 10 + c - '0';
              c = br.read();
          }
          return ril(x);
      }
  
      long[] rkll() throws IOException {
          int sign = 1;
          int c = br.read();
          int x = 0;
          if (c == '-') {
              sign = -1;
              c = br.read();
          }
          while (c >= '0' && c <= '9') {
              x = x * 10 + c - '0';
              c = br.read();
          }
          return rll(x);
      }
  
      char[] rs() throws IOException {
          return br.readLine().toCharArray();
      }
  
      void sort(int[] A) {
          Random r = new Random();
          for (int i = A.length-1; i > 0; i--) {
              int j = r.nextInt(i+1);
              int temp = A[i];
              A[i] = A[j];
              A[j] = temp;
          }
          Arrays.sort(A);
      }
  
      void printDouble(double d) {
          pw.printf("%.16f", d);
      }
  }