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;
  int INF = 0x3f3f3f3f;
  long long LINF = 0x3f3f3f3f3f3f3f3f;
  int dy[]{-1, 1, 0, 0};
  int dx[]{0, 0, -1, 1};
  int k;
  string n;
  string solve1() {
    string ret = string(n.length(), '9');
    for (char a = '8'; a >= '0'; a--) {
      string t = string(n.length(), a);
      if (t >= n) ret = t;
    }
    return ret;
  }
  string solve2() {
    string ret = solve1();
    for (char a = '0'; a <= '8'; a++) {
      for (char b = a + 1; b <= '9'; b++) {
        bool flag = true;
        for (int i = (0); i < (n.length()); ++i) {
          if (n[i] < b) {
            string t = n;
            if (t[i] < a)
              t[i] = a;
            else
              t[i] = b;
            for (int j = (i + 1); j < (n.length()); ++j) t[j] = a;
            if (t < ret) ret = t;
          }
          if (n[i] != a && n[i] != b) {
            flag = false;
            break;
          }
        }
        if (flag) {
          return n;
        }
      }
    }
    return ret;
  }
  string solve3() {
    while (1) {
      set<char> s;
      for (auto& e : n) {
        s.insert(e);
      }
      if (s.size() <= k) return n;
      s.clear();
      for (int i = 0;; i++) {
        s.insert(n[i]);
        if (s.size() > k) {
          int p = i;
          while (n[p] == '9') --p;
          n[p]++;
          for (int j = (p + 1); j < (n.length()); ++j) n[j] = '0';
          break;
        }
      }
    }
  }
  void solve() {
    cin >> n >> k;
    if (k == 1)
      cout << solve1() << '\n';
    else if (k == 2)
      cout << solve2() << '\n';
    else
      cout << solve3() << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    int TC;
    cin >> TC;
    while (TC--) {
      solve();
    }
    return 0;
  }
  