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.*;
  import java.lang.*;
  import java.io.*;
  
  public class F {
  
  	public static void main(String[] args) throws Exception {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		int t = Integer.parseInt(br.readLine());
  		StringBuilder sb = new StringBuilder();
  		while (t-- > 0) {
  			String[] scn = (br.readLine()).trim().split(" ");
  			String str = scn[0];
  			int k = Integer.parseInt(scn[1]);
  			int n = str.length();
  			int[] orig = new int[n];
  			for (int i = 0; i < n; i++) {
  				orig[i] = str.charAt(i) - '0';
  			}
  			int[] myans = new int[n];
  			fun(myans, orig, 0, n, k, 0);
  			for (int ele : myans) {
  				sb.append(ele);
  			}
  			sb.append("\n");
  		}
  		System.out.println(sb);
  		return;
  
  	}
  
  	public static int fun(int[] myans, int[] orig, int i, int n, int k, int myk) {
  		if (myk > k) {
  			return -1;
  		}
  		if (!check(myans, orig, i)) {
  			return -1;
  		}
  		if (i >= n) {
  			return 1;
  		}
  		for (int dig = 0; dig < 10; dig++) {
  			if (i == 0 && dig == 0) {
  				continue;
  			}
  			myans[i] = dig;
  			boolean present = false;
  			for (int j = 0; j < i; j++) {
  				if (myans[j] == dig)
  					present = true;
  			}
  			int recans = fun(myans, orig, i + 1, n, k, myk + (present ? 0 : 1));
  			if (recans != -1) {
  				return 1;
  			}
  		}
  		return -1;
  
  	}
  
  	public static boolean check(int[] myans, int[] orig, int n) {
  		boolean equal = false;
  		for (int i = 0; i < n; i++) {
  			if (myans[i] > orig[i]) {
  				return true;
  			}
  			if ((myans[i] < orig[i])) {
  				return false;
  			}
  		}
  		return true;
  
  	}
  
  	public static void sort(long[] arr) {
  		int n = arr.length;
  		for (int i = 0; i < n; i++) {
  			int idx = (int) (Math.random() * n);
  			long temp = arr[i];
  			arr[i] = arr[idx];
  			arr[idx] = temp;
  		}
  		Arrays.sort(arr);
  	}
  
  	public static void sort(int[] arr) {
  		int n = arr.length;
  		for (int i = 0; i < n; i++) {
  			int idx = (int) (Math.random() * n);
  			int temp = arr[i];
  			arr[i] = arr[idx];
  			arr[idx] = temp;
  		}
  		Arrays.sort(arr);
  	}
  
  	public static void print(long[][] dp) {
  		for (long[] a : dp) {
  			for (long ele : a) {
  				System.out.print(ele + " ");
  			}
  			System.out.println();
  		}
  	}
  
  	public static void print(int[][] dp) {
  		for (int[] a : dp) {
  			for (int ele : a) {
  				System.out.print(ele + " ");
  			}
  			System.out.println();
  		}
  	}
  
  	public static void print(int[] dp) {
  		for (int ele : dp) {
  			System.out.print(ele + " ");
  		}
  		System.out.println();
  	}
  
  	public static void print(long[] dp) {
  		for (long ele : dp) {
  			System.out.print(ele + " ");
  		}
  		System.out.println();
  	}
  
  }
  