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;
  using ull = unsigned long long;
  using ll = long long;
  const int N = 200005, K = 10, MOD = 1e9 + 7;
  int t, n, k;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> t;
    while (t--) {
      cin >> n >> k;
      string s = to_string(n);
      int len = s.length();
      ll val = 0;
      if (k == 1) {
        int d = s[0] - '0';
        if (d == 9)
          val = 1000000005;
        else {
          ++d;
          for (int i = 0; i < len; ++i) {
            val = val * 10 + d;
          }
        }
      } else {
        int d = s[0] - '0';
        if (d == 9)
          val = 1000000005;
        else {
          ++d;
          for (int i = 0; i < len; ++i) {
            val = val * 10 + (i == 0 ? d : 0);
          }
        }
      }
      for (int i = 0; i < len; ++i) {
        if (i != len - 1 && s[i + 1] == '9') continue;
        ll vl2 = 0;
        vector<char> dv(10, 0);
        for (int j = 0; j <= i; ++j) {
          dv[s[j] - '0'] = 1;
        }
        int cnt = 0;
        for (int j = 0; j < 10; ++j) {
          cnt += dv[j];
        }
        if (cnt > k) break;
        if (i == len - 1) {
          val = n;
          break;
        }
        if (cnt == k) {
          for (int j = 0; j <= i; ++j) {
            vl2 = vl2 * 10 + (s[j] - '0');
          }
          int mn = 0;
          while (!dv[mn]) ++mn;
          int lg = s[i + 1] - '0' + 1;
          while (lg < 10 && !dv[lg]) ++lg;
          if (lg == 10) continue;
          vl2 = vl2 * 10 + lg;
          for (int j = i + 2; j < len; ++j) {
            vl2 = vl2 * 10 + mn;
          }
          val = min(val, vl2);
        } else {
          for (int j = 0; j <= i; ++j) {
            vl2 = vl2 * 10 + (s[j] - '0');
          }
          vl2 = vl2 * 10 + s[i + 1] - '0' + 1;
          int mn = 0;
          if (dv[s[i + 1] - '0' + 1] || cnt < k - 1) {
            mn = 0;
          } else {
            dv[s[i + 1] - '0' + 1] = 1;
            while (!dv[mn]) ++mn;
          }
          for (int j = i + 2; j < len; ++j) {
            vl2 = vl2 * 10 + mn;
          }
          val = min(val, vl2);
        }
      }
      cout << val << '\n';
    }
    return 0;
  }
  