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>
  #pragma GCC optimize("O3")
  using namespace std;
  void solve() {
    int n, k;
    cin >> n >> k;
    if (n == 1) {
      cout << 1 << '\n';
      return;
    }
    n--;
    string s = to_string(n);
    string ans;
    for (int i = s.size() - 1; i >= 0; i--) {
      int dig = s[i] - '0';
      for (int j = dig + 1; j <= 9; j++) {
        set<int> gg;
        gg.insert(j);
        for (int x = 0; x < i; x++) gg.insert(s[x] - '0');
        if (gg.size() > k) continue;
        if ((int)gg.size() < k) {
          cout << s.substr(0, i) + (char)('0' + j) + string(s.size() - 1 - i, '0')
               << '\n';
          return;
        } else {
          int min_dig = 9;
          min_dig = min(min_dig, j);
          for (int p = 0; p < i; p++) min_dig = min(min_dig, s[p] - '0');
          cout << s.substr(0, i) + (char)('0' + j) +
                      string(s.size() - 1 - i, min_dig + '0')
               << '\n';
          return;
        }
      }
    }
    if (k == 1) {
      cout << string(s.size() + 1, '1') << '\n';
    } else {
      cout << '1' + string(s.size(), '0') << '\n';
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int tst;
    cin >> tst;
    while (tst--) {
      solve();
    }
    return 0;
  }
  