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.util.*;
  import java.io.*;
  
  public class c {
  
      public static void main(String[] args){
  
          FastScanner sc = new FastScanner();
          int t = sc.nextInt();
          while(t-- > 0){
  
              int n = sc.nextInt();
              long k = sc.nextLong();
              k++;
  
              long arr[] = new long[n];
              for(int i=0; i<n; i++){
  
                  arr[i] = (long)Math.pow(10, sc.nextLong());
  
              }
  
              long currentNum = 0;
              long notes = 0;
  
              for(int i=0; i<n-1; i++){
  
                  if(notes == k) break;
  
                  if(k - notes >= arr[i+1]/arr[i]-1){
                      notes += arr[i+1]/arr[i]-1;
                      currentNum += arr[i] * (arr[i+1]/arr[i]-1);
                  }
                  else{
                      currentNum += (k-notes)*arr[i];
                      notes += k-notes;
                  }
  
                 // System.out.println("Notes: " + notes + " CurrentNUm: " + currentNum);
  
              }
  
              if(notes < k){
  
                  currentNum += (k-notes) * arr[arr.length-1];
  
              }
  
              System.out.println(currentNum);
  
          }
  
      }
  
  }
  
  class FastScanner {
  
      //I don't understand how this works lmao
      private int BS = 1 << 16;
      private char NC = (char) 0;
      private byte[] buf = new byte[BS];
      private int bId = 0, size = 0;
      private char c = NC;
      private double cnt = 1;
      private BufferedInputStream in;
  
      public FastScanner() {
          in = new BufferedInputStream(System.in, BS);
      }
  
      public FastScanner(String s) {
          try {
              in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
          } catch (Exception e) {
              in = new BufferedInputStream(System.in, BS);
          }
      }
  
      private char getChar() {
          while (bId == size) {
              try {
                  size = in.read(buf);
              } catch (Exception e) {
                  return NC;
              }
              if (size == -1) return NC;
              bId = 0;
          }
          return (char) buf[bId++];
      }
  
      public int nextInt() {
          return (int) nextLong();
      }
  
      public int[] nextInts(int N) {
          int[] res = new int[N];
          for (int i = 0; i < N; i++) {
              res[i] = (int) nextLong();
          }
          return res;
      }
  
      public long[] nextLongs(int N) {
          long[] res = new long[N];
          for (int i = 0; i < N; i++) {
              res[i] = nextLong();
          }
          return res;
      }
  
      public long nextLong() {
          cnt = 1;
          boolean neg = false;
          if (c == NC) c = getChar();
          for (; (c < '0' || c > '9'); c = getChar()) {
              if (c == '-') neg = true;
          }
          long res = 0;
          for (; c >= '0' && c <= '9'; c = getChar()) {
              res = (res << 3) + (res << 1) + c - '0';
              cnt *= 10;
          }
          return neg ? -res : res;
      }
  
      public double nextDouble() {
          double cur = nextLong();
          return c != '.' ? cur : cur + nextLong() / cnt;
      }
  
      public double[] nextDoubles(int N) {
          double[] res = new double[N];
          for (int i = 0; i < N; i++) {
              res[i] = nextDouble();
          }
          return res;
      }
  
      public String next() {
          StringBuilder res = new StringBuilder();
          while (c <= 32) c = getChar();
          while (c > 32) {
              res.append(c);
              c = getChar();
          }
          return res.toString();
      }
  
      public String nextLine() {
          StringBuilder res = new StringBuilder();
          while (c <= 32) c = getChar();
          while (c != '\n') {
              res.append(c);
              c = getChar();
          }
          return res.toString();
      }
  
      public boolean hasNext() {
          if (c > 32) return true;
          while (true) {
              c = getChar();
              if (c == NC) return false;
              else if (c > 32) return true;
          }
      }
  }