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.Scanner;
  public class Main
  {
      static String go(int i,int n,int dist,String nw,int k,String ori){
          if(dist>k){
              return "";
          }
          if((nw.substring(0,i)).compareTo(ori.substring(0,i))<0){
              return "";
          }
          if(i>=n){
              return nw;
          }
          for(char val='0';val<='9';++val){
              nw = nw.substring(0,i)+val+nw.substring(i+1);//replacing val at index i in string nw
              boolean found=false;
              for(int j=0;j<i;++j){
                  if(nw.charAt(j)==val){
                      found=true;
                      break;
                  }
              }
              String chk;
              if(found==false){
                  chk=go(i+1,n,dist+1,nw,k,ori);
              }else{
                  chk=go(i+1,n,dist,nw,k,ori);
              }
              if(chk.length()!=0){
                  return chk;
              }
          }
          return "";
      }
  	public static void main(String[] args) {
  		Scanner sc=new Scanner(System.in);
  		int test=sc.nextInt();
  		sc.nextLine();
  		while((test--)>0){
  		    int n=sc.nextInt();
  		    int k=sc.nextInt();
  		    String s=String.valueOf(n);
  		    String str="";
  		    for(int i=0;i<s.length();i++){
  		        str=str+"0";
  		    }
  		    String ans=go(0,s.length(),0,str,k,s);
  		    System.out.println(ans);
  		}
  	}
  }
  
  
  