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;
  void SolveF();
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    SolveF();
    return 0;
  }
  void SolveF() {
    int T;
    cin >> T;
    while (T--) {
      long long n;
      int k;
      cin >> n >> k;
      int currk[11];
      int minInt[11];
      int a[10];
      memset(a, 0, sizeof(a));
      string s = to_string(n);
      int lens = s.length();
      ++a[s[0] - '0'];
      currk[0] = 1;
      minInt[0] = s[0] - '0';
      for (int i = 1; i < lens; ++i) {
        if (a[s[i] - '0'] == 0) {
          currk[i] = currk[i - 1] + 1;
          minInt[i] =
              ((minInt[i - 1]) < (s[i] - '0') ? (minInt[i - 1]) : (s[i] - '0'));
        } else {
          currk[i] = currk[i - 1];
          minInt[i] = minInt[i - 1];
        }
        ++a[s[i] - '0'];
      }
      if (k == currk[lens - 1]) {
        cout << s << endl;
        continue;
      } else if (k > lens) {
        cout << s << endl;
      } else if (k > currk[lens - 1]) {
        cout << s << endl;
      } else {
        while (1) {
          int pos = 0;
          for (int i = 0; i < 10; ++i) {
            a[i] = 0;
            currk[i] = 0;
          }
          currk[0] = 1;
          a[s[0] - '0'] = 1;
          for (int i = 1; i < lens; ++i) {
            if (a[s[i] - '0'] == 0)
              currk[i] = currk[i - 1] + 1;
            else
              currk[i] = currk[i - 1];
            ++a[s[i] - '0'];
          }
          if (k >= currk[lens - 1]) break;
          for (int i = 0; i < lens; ++i) {
            if (currk[i] == k + 1) {
              pos = i;
              break;
            }
          }
          while (s[pos] == 9 + '0') --pos;
          ++s[pos];
          for (int i = pos + 1; i < lens; ++i) s[i] = '0';
        }
        cout << s << endl;
      }
    }
  }
  