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;
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  string get(long long i, long long n, long long dis, long long k, string org,
             string cur) {
    if (dis > k) {
      return "";
    }
    if (cur.substr(0, i) < org.substr(0, i)) {
      return "";
    }
    if (i >= n) {
      return cur;
    }
    for (char x = '0'; x <= '9'; x += 1) {
      cur[i] = x;
      bool check = false;
      for (long long j = 0; j < i; j++) {
        if (cur[j] == x) {
          check = true;
          break;
        }
      }
      string lol = get(i + 1, n, dis + (!check), k, org, cur);
      if (lol.size()) {
        return lol;
      }
    }
    return "";
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    string s = to_string(n);
    string ans = get(0, s.size(), 0, k, s, string(s.size(), '0'));
    cout << ans << endl;
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  