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 long long oo = INT_MAX;
  long long t, n, k, cnt, a[100005], b[100005], fr[100005];
  long long build(long long poz) {
    long long auxcnt = cnt;
    if (cnt > k) {
      return oo;
    }
    for (long long i = poz + 1; i <= a[0]; i++) {
      b[i] = -1;
      for (long long c = 0; c <= 9; c++) {
        if (i == poz + 1 && c <= a[i]) {
          continue;
        }
        if (fr[c] || (!fr[c] && cnt < k)) {
          ++fr[c];
          b[i] = c;
          if (fr[c] == 1) {
            ++cnt;
          }
          break;
        }
      }
    }
    for (long long i = poz + 1; i <= a[0]; i++) {
      if (b[i] == -1) {
        continue;
      }
      --fr[b[i]];
    }
    cnt = auxcnt;
    long long val = 0;
    for (long long i = 1; i <= poz; i++) {
      val = val * 10 + a[i];
    }
    for (long long i = poz + 1; i <= a[0]; i++) {
      if (b[i] == -1) {
        return oo;
      }
      val = val * 10 + b[i];
    }
    return val;
  }
  void solve_test() {
    cin >> n >> k;
    for (long long i = 0; i <= 9; i++) {
      fr[i] = 0;
    }
    cnt = 0;
    long long aux = n;
    a[0] = 0;
    while (n) {
      a[++a[0]] = n % 10;
      b[a[0]] = n % 10;
      n /= 10;
    }
    reverse(a + 1, a + a[0] + 1);
    long long Min = build(0);
    for (long long i = 1; i <= a[0]; i++) {
      ++fr[a[i]];
      if (fr[a[i]] == 1) {
        ++cnt;
      }
      Min = min(Min, build(i));
    }
    cout << Min << '\n';
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> t;
    for (long long test = 1; test <= t; test++) {
      solve_test();
    }
    return 0;
  }
  