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>
  using namespace std;
  long long t, k, n, ans[21], cnt[21];
  string s;
  bool dfs(long long now, long long val, bool tag) {
    if (val > k) return 0;
    if (now == n && val <= k) {
      for (register long long i = 0; i < n; ++i) cout << ans[i];
      cout << endl;
      return 1;
    }
    for (register long long i = tag ? s[now] - '0' : 0; i <= 9; ++i) {
      ++cnt[i];
      ans[now] = i;
      if (cnt[i] == 1) {
        if (dfs(now + 1, val + 1, tag && i == s[now] - '0')) return 1;
      } else if (dfs(now + 1, val, tag && i == s[now] - '0'))
        return 1;
      --cnt[i];
    }
    return 0;
  }
  signed main() {
    cin >> t;
    while (t--) {
      for (register long long i = 0; i <= 10; ++i) cnt[i] = 0;
      for (register long long i = 0; i <= 10; ++i) ans[i] = 0;
      cin >> s >> k;
      n = s.length();
      if (!dfs(0, 0, 1)) {
        cout << 10;
        for (register long long i = 2; i < k; ++i) cout << i;
        cout << endl;
      }
    }
    return 0;
  }
  