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.io.*;
  import java.util.*;
   
  public class NearestBeautifulNumberHardVersion {
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static StringTokenizer st;
  	
  	public static void main(String[] args) throws IOException {
  		int t = readInt();
  		
  		while (t-- > 0) {
  			int n = readInt(), k = readInt();
  			char[] a = String.valueOf(n).toCharArray();
  			System.out.println(solve(a, k));
  		}
  	}
  	
  	static String solve(char[] a, int k) {
  		HashSet<Character> s = new HashSet<Character>();
  		for (char c: a) s.add(c);
  		if (s.size() <= k) {
  			String str = "";
  			for (char c: a) str += c;
  			return str;
  		}
  		
  		s.clear();
  		int ptr = 0;
  		for (; ; ptr ++) {
  			s.add(a[ptr]);
  			if (s.size() > k) {
  				if (a[ptr] == '9') ptr --;
  				a[ptr] ++;
  				for (int i = ptr + 1; i < a.length; i ++) a[i] = '0';
  				return solve(a, k);
  			}
  		}
  	}
   
  	static String next() throws IOException {
  		while (st == null || !st.hasMoreTokens()) 
  			st = new StringTokenizer(br.readLine().trim());
  		return st.nextToken();
  	}
  	static long readLong() throws IOException {
  		return Long.parseLong(next());
  	}
  	static int readInt() throws IOException {
  		return Integer.parseInt(next());
  	}
  	static double readDouble() throws IOException {
  		return Double.parseDouble(next());
  	}
  	static char readCharacter() throws IOException {
  		return next().charAt(0);
  	}
  	static String readLine() throws IOException {
  		return br.readLine().trim();
  	}
  }