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.lang.*;
  import java.io.*;
   
  public class snackDown3
  {
   public static class FastReader {
   
  		BufferedReader b;
  		StringTokenizer s;
  		public FastReader() {
  			b=new BufferedReader(new InputStreamReader(System.in));
  		}
  		String next() {
  			while(s==null ||!s.hasMoreElements()) {
  				try {
  					s=new StringTokenizer(b.readLine());
  				}
  				catch(IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return s.nextToken();
  		}
  		public int nextInt() {
  			return Integer.parseInt(next());
  		}
  		public long nextLong() {
  			return Long.parseLong(next());
  		}
  		public double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		String nextLine() {
  			String str="";
  			try {
  				str=b.readLine();
  			}
  			catch(IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  		boolean hasNext() {
  		    if (s != null && s.hasMoreTokens()) {
  		        return true;
  		    }
  		    String tmp;
  		    try {
  		        b.mark(1000);
  		        tmp = b.readLine();
  		        if (tmp == null) {
  		            return false;
  		        }
  		        b.reset();
  		    } catch (IOException e) {
  		        return false;
  		    }
  		    return true;
  		}
  	
  }
   
  	public static void main (String[] args) throws Exception
  	{
  		// your code goes here
  		FastReader sc=new FastReader();
  		int t=sc.nextInt();
          BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
  		while(t--!=0) {
  		   int n=sc.nextInt();
  		   long k=sc.nextLong();
  		   int a[]=new int[n];
  		   long start=0;
  //		   int p=(int)Math.pow(10, 9);
  		   for(int i=0;i<n;i++) {
  			   a[i]=sc.nextInt();
  		   }
  		   k++;
  		   int cnt=1;
  		   for(int i=0;i<n-1;i++) {
  			   long num=(long)Math.pow(10,a[i]);
  			   int diff=a[i+1]-a[i];
  			   long numm=((long)Math.pow(10, diff))-1;
  			   if(k>numm) {
  				   k-=numm;
  				   start+=(numm*num);
  			   }
  			   else if(k<=numm) {
  				   
  				   start=(k*num)+start;
  				   k=0;
  				   break;
  			   }  
  		   }
  		   if(k>0) {
  			   long num=((long)Math.pow(10, a[n-1]));
  			  start=((k)*num)+start;
  		   }
  		    log.write(start+"");
  			log.write("\n");
  			log.flush();
  		}
  		
  	}
   
   
  }