description: It is a complicated version of problem F1. The difference between them is the constraints (F1: k ≤ 2, F2: k ≤ 10).
  
  You are given an integer n. Find the minimum integer x such that x ≥ n and the number x is k-beautiful.
  
  A number is called k-beautiful if its decimal representation having no leading zeroes contains no more than k different digits. E.g. if k = 2, the numbers 3434443, 55550, 777 and 21 are k-beautiful whereas the numbers 120, 445435 and 998244353 are not.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
  
  Each test case consists of one line containing two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 10).
  
  Output
  
  For each test case output on a separate line x — the minimum k-beautiful integer such that x ≥ n.
  
  Example
  
  Input
  
  
  6
  2021 3
  177890 2
  34512 3
  724533 4
  998244353 1
  12345678 10
  
  
  Output
  
  
  2021
  181111
  34533
  724542
  999999999
  12345678
solution: import java.util.*;
  
  public class B {
  
  	public static void main(String[] args) {
  
  		Scanner scan=new Scanner(System.in);
  	
  		int t=scan.nextInt();
  		while(t-->0) {
  			String n=scan.next();
  			int k=scan.nextInt();
  			set=new HashSet<Character>();
  			System.out.println(helper(n, k, ""));
  			
  		}
  	
  	}
  	private static char[] nums= {'0','1','2','3','4','5','6','7','8','9'};
  	private static HashSet<Character> set;  
  	private static String helper(String n, int k,String prev){
  		
  		if(prev.length()!=0 && Long.valueOf(prev)>= Long.valueOf(n)) return prev;
  		
  //		for(int i=0;i<n.length()-prev.length()+1;i++) {
  			for(char c:nums) {
  				String tmp=prev+c;
  				boolean b=false;
  				if(!set.contains(c)) {
  					set.add(c);
  					b=true;
  				}
  					
  //				System.out.println(tmp+"-cmpare-"+(n.substring(0,Math.min(tmp.length(),n.length())))+", "+
  //						(Long.valueOf(tmp)>= Long.valueOf(n.substring(0,Math.min(tmp.length(),n.length()))) && set.size()<=k));/////////
  				if(Long.valueOf(tmp)>= Long.valueOf(n.substring(0,Math.min(tmp.length(),n.length()))) && set.size()<=k) {
  					String newn= helper(n,k,tmp);
  					if(Long.valueOf(newn)>= Long.valueOf(n))return newn;
  				}
  				if(b) set.remove(c);
  			}
  			
  			
  //		}
  		return "0";
  	}
  }
  