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: #include <bits/stdc++.h>
  int funcK(int v) {
    int num[10];
    for (int i = 0; i < 10; ++i) num[i] = 0;
    while (v > 0) {
      num[v % 10] = 1;
      v /= 10;
    }
    int x = 0;
    for (int i = 0; i < 10; ++i) x += num[i];
    return x;
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      int n, k;
      scanf("%d%d", &n, &k);
      while (funcK(n) > k) {
        int o = n;
        while (funcK(o) > k) o /= 10;
        int p = 1;
        while (n / p != o) p *= 10;
        p /= 10;
        n = (n / p) * p + p;
      }
      printf("%d\n", n);
    }
  }
  