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){
          JS scan = new JS();
          PrintWriter out = new PrintWriter(System.out);
          long[] powten = new long[12];
          powten[0] = 1;
          for(int i = 1; i < powten.length; i++) powten[i] = powten[i-1]*10;
          int t = scan.nextInt();
          Q: for(int q = 1; q <= t; q++){
              int n = scan.nextInt();
              long k = scan.nextLong()+1;
              long[] a = new long[n];
              for(int i = 0; i < n; i++) a[i] = scan.nextLong();
  
              long[] diff = new long[n-1];
              for(int i = 0; i < n-1; i++) diff[i] = (powten[(int)a[i+1]]-powten[(int)a[i]])/powten[(int)a[i]];
  
              long pre = 0;
  
              long num = 0;
              for(int i = 0; i < n-1; i++){
                  if(pre+diff[i] >= k){
                      // out.println(pre);
                      // out.println(i+" here");
                      long d = k-pre;
                      num += d*powten[(int)a[i]];
                      out.println(num);
                      continue Q;
                  }
                  pre += diff[i];
                  num += diff[i]*powten[(int)a[i]];
              }
              long d = k-pre;
              num += d*powten[(int)a[n-1]];
              out.println(num);
          }
          out.flush();
      }
  
      static class JS {
          public int BS = 1<<16;
          public char NC = (char)0;
          byte[] buf = new byte[BS];
          int bId = 0, size = 0;
          char c = NC;
          double num = 1;
          BufferedInputStream in;
   
          public JS() {
              in = new BufferedInputStream(System.in, BS);
          }
   
          public JS(String s) throws FileNotFoundException {
              in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
          }
   
          public char nextChar(){
              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 long nextLong() {
              num=1;
              boolean neg = false;
              if(c==NC)c=nextChar();
              for(;(c<'0' || c>'9'); c = nextChar()) {
                  if(c=='-')neg=true;
              }
              long res = 0;
              for(; c>='0' && c <='9'; c=nextChar()) {
                  res = (res<<3)+(res<<1)+c-'0';
                  num*=10;
              }
              return neg?-res:res;
          }
   
          public double nextDouble() {
              double cur = nextLong();
              return c!='.' ? cur:cur+(cur < 0 ? -1*nextLong()/num : nextLong()/num);
          }
   
          public String next() {
              StringBuilder res = new StringBuilder();
              while(c<=32)c=nextChar();
              while(c>32) {
                  res.append(c);
                  c=nextChar();
              }
              return res.toString();
          }
      }
  }