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;
  const int maxn = 2e5 + 7;
  const int maxm = 26e3 + 7;
  const long long mod = 998244353;
  int n, k;
  int flag[12];
  vector<int> v;
  string s;
  int check(int pos) {
    set<int> second;
    for (int i = (0); i <= (pos); ++i) {
      second.insert(s[i] - '0');
    }
    return (second.size() < k);
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      cin >> s >> k;
      memset(flag, 0, sizeof(flag));
      v.clear();
      int pos = -1;
      for (int i = (0), __ = (s.size()); i < __; ++i) {
        if (!flag[s[i] - '0']) {
          if (v.size() == k) {
            pos = i;
            break;
          } else
            flag[s[i] - '0']++, v.push_back(s[i] - '0');
        }
      }
      if (pos == -1) {
        cout << s << '\n';
        continue;
      }
      sort(v.begin(), v.end());
      int tb = 0;
      for (int i = pos; i >= 0; i--) {
        int p = upper_bound(v.begin(), v.end(), s[i] - '0') - v.begin();
        if (check(i - 1) && s[i] != '9') {
          int fd = 0;
          for (int j = 0; j < v.size(); j++) {
            if (s[i] - '0' == v[j])
              v.erase(v.begin() + j), j--;
            else if (s[i] - '0' + 1 == v[j])
              fd = 1;
          }
          if (!fd) {
            v.push_back(s[i] - '0' + 1);
            sort(v.begin(), v.end());
          }
          s[i] = char(s[i] + 1);
          pos = i;
          break;
        } else if (p != v.size()) {
          s[i] = char(v[p] + '0');
          pos = i;
          break;
        }
        if (!i) tb = 1;
      }
      if (v.size() < k) v[0] = 0;
      if (tb) {
        if (k == 1) {
          for (int i = (1); i <= (s.size() + 1); ++i) cout << 1;
        } else {
          cout << 1;
          for (int i = (1); i <= (s.size()); ++i) cout << 0;
        }
      } else {
        for (int i = (0); i <= (pos); ++i) cout << s[i];
        for (int i = (pos + 1), __ = (s.size()); i < __; ++i) cout << v[0];
      }
      cout << '\n';
    }
  }
  