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 fun(long long n, long long k) {
    string s = to_string(n);
    long long sz = s.size();
    vector<long long> mini(sz);
    long long minimum = s[0] - '0';
    set<char> st;
    vector<long long> f(sz);
    for (long long i = 0; i < sz; i++) {
      mini[i] = min(minimum, (long long)(s[i] - '0'));
      st.insert(s[i]);
      f[i] = st.size();
    }
    if (st.size() <= k) return n;
    long long x = 10000000000007;
    long long c = 0;
    for (long long i = sz - 2; i >= 0; i--) {
      c++;
      if (f[i] == k) {
        set<long long> digits;
        for (long long j = 0; j <= i; j++) {
          st.insert(s[j] - '0');
        }
        long long d = s[i + 1] - '0';
        if (digits.lower_bound(d) != digits.end()) {
          d = *(digits.lower_bound(d));
          for (long long j = i + 1; j < sz; j++) {
            s[j] = d + '0';
          }
        } else {
          long long z = pow(10, c - 1);
          n = n / z;
          n++;
          n *= z;
          long long p = fun(n, k);
          return p;
        }
      }
    }
    return -1;
  }
  void solve(long long cc) {
    long long n, k;
    cin >> n >> k;
    long long x = fun(n, k);
    string s1 = to_string(x);
    string s2 = to_string(n);
    cout << x;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ;
    long long t = 1;
    cin >> t;
    long long c = 1;
    while (t--) {
      ios_base::sync_with_stdio(false);
      cin.tie(0);
      cout.tie(0);
      ;
      solve(c);
      c++;
      cout << endl;
    }
    return 0;
  }
  