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 INF = (int)(1e9) + 10;
  int n, k;
  int ans;
  string s;
  void update(int pos, int mask, int num) {
    if (pos == (int)s.size()) {
      ans = num;
      return;
    }
    if (__builtin_popcount(mask) < k) {
      for (int i = pos; i < (int)s.size(); i++) num *= 10;
      ans = num;
      return;
    }
    if (__builtin_popcount(mask) == k) {
      for (int d = 0; d < 10; d++) {
        if (((1 << d) | mask) == mask) {
          for (int i = pos; i < (int)s.size(); i++) num = num * 10 + d;
          ans = num;
          return;
        };
      }
    };
    return;
  }
  bool go(int pos, bool upper, int mask, int num) {
    if ((pos == (int)s.size()) || upper) {
      update(pos, mask, num);
      return true;
    }
    int low_d = (upper) ? 0 : s[pos] - '0';
    for (int d = low_d; d < 10; d++) {
      int new_mask = mask | (1 << d);
      if (__builtin_popcount(new_mask) <= k) {
        if (go(pos + 1, upper || (d > s[pos] - '0'), new_mask, num * 10 + d))
          return true;
      }
    }
    return false;
  };
  void solve() {
    cin >> n >> k;
    s = to_string(n);
    ans = 0;
    go(0, 0, 0, 0);
    cout << ans << '\n';
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int T = 1;
    cin >> T;
    while (T--) solve();
    return 0;
  }
  