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();
  		
  		outer: while (t-- > 0) {
  			int n = readInt(), k = readInt();
  			String s = String.valueOf(n);
  			TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
  			
  			for (int i = 0; i < s.length(); i ++) {
  				int x = Character.getNumericValue(s.charAt(i));
  				if (!map.containsKey(x) && map.size() == k) {
  					if (map.ceilingKey(x) == null) {
  						int idx = -1, val = 0, mx = map.lastKey(), min = map.firstKey();
  						for (int j = i - 1; j >= 0; j --) {
  							int d = Character.getNumericValue(s.charAt(j));
  							if (d != mx && j > map.get(mx)) { val = d; idx = j; break; }
  						}
  						if (idx == -1) {
  							if (k == 1) for (int j = 0; j < s.length(); j ++) System.out.print(mx + 1);
  							else {
  								for (int j = 0; j < map.get(mx); j ++) System.out.print(s.charAt(j));
  								System.out.print(mx + 1);
  								for (int j = map.get(mx) + 1; j < s.length(); j ++) System.out.print(min);
  							}
  						}
  						else {
  							if (map.get(val) != idx) {
  								for (int j = 0; j < idx; j ++) System.out.print(s.charAt(j));
  								System.out.print(map.higherKey(val));
  								for (int j = idx + 1; j < s.length(); j ++) System.out.print(min);
  							} else {
  								for (int j = 0; j < idx; j ++) System.out.print(s.charAt(j));
  								System.out.print(val + 1);
  								if (!map.containsKey(val + 1)) {
  									if (min == val) for (int j = idx + 1; j < s.length(); j ++) System.out.print(val + 1);
  									else for (int j = idx + 1; j < s.length(); j ++) System.out.print(min);
  								} else {
  									for (int j = idx + 1; j < s.length(); j ++) System.out.print(0);
  								}
   							}
  						}
  					} else {
  						int c = map.ceilingKey(x), min = map.firstKey();
  						for (int j = 0; j < i; j ++) System.out.print(s.charAt(j));
  						System.out.print(c);
  						for (int j = i + 1; j < s.length(); j ++) System.out.print(min);
  					}
  					System.out.println();
  					continue outer;
  				} else if (!map.containsKey(x)) map.put(x, i);
  			}
  			System.out.println(n);
  		}
  	}
   
  	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();
  	}
  }