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>
  #pragma GCC optimize("unroll-loops")
  #pragma GCC optimize("O3")
  #pragma GCC target("avx2")
  using namespace std;
  const long long N = 1e5 + 1, mod = 1e9 + 7;
  long long itos(long long n, long long ch[]) {
    long long j = n, l = 0;
    while (j) {
      j /= 10;
      l++;
    }
    for (long long i = l - 1; i >= 0; i -= 1) {
      ch[i] = n % 10;
      n /= 10;
    }
    return l;
  }
  long long stoi(long long ch[], long long l) {
    long long n = 0;
    for (long long i = 0; i < l; i += 1) {
      n *= 10;
      n += ch[i];
      ;
    }
    return n;
  }
  void solve() {
    string n;
    long long k;
    cin >> n >> k;
    while (true) {
      set<char> s;
      for (auto x : n) {
        s.insert(x);
      }
      if (s.size() <= k) {
        cout << n << "\n";
        return;
      }
      s.clear();
      for (long long i = 0; i < n.size(); i += 1) {
        s.insert(n[i]);
        if (s.size() > k) {
          while (n[i] == '9') i--;
          n[i]++;
          i++;
          for (; i < n.size(); i += 1) {
            n[i] = '0';
          }
        }
      }
    }
    return;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  