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 S = int(2e5) + 100;
  const int INF = int(1e9) + 7;
  const long long MOD = long long(1e9) + 7;
  const double PI = 3.141592653589793238;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <typename T>
  void print(T& v) {
    for (auto& x : v) cout << x << " ";
    cout << endl;
  }
  long long ans(long long n, int k) {
    vector<int> dg;
    long long on = n;
    while (n) {
      dg.emplace_back(n % 10);
      n /= 10;
    }
    reverse(dg.begin(), dg.end());
    int i = 0, seen[10] = {}, cnt = 0;
    for (; i < (int)(dg).size(); i++) {
      if (!seen[dg[i]]) {
        cnt++;
        seen[dg[i]] = 1;
      }
      if (cnt > k) break;
    }
    if (cnt <= k) return on;
    long long m = 0;
    for (int j = 0; j <= i; j++) m = m * 10 + dg[j];
    m++;
    for (int j = (0); j < ((int)(dg).size() - i - 1); ++j) m *= 10;
    return ans(m, k);
  }
  int t, k;
  long long n;
  int main() {
    cin.sync_with_stdio(0);
    cin.tie(0);
    cin >> t;
    while (t--) {
      cin >> n >> k;
      cout << ans(n, k) << endl;
    }
    return 0;
  }
  