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 mod = 1000000007;
  const unsigned gen_seed =
      std::chrono::system_clock::now().time_since_epoch().count();
  std::mt19937_64 gen(gen_seed);
  int test(int n) {
    std::vector<int> cnt(10, 0);
    int ret = 0;
    while (n > 0) {
      int k = n % 10;
      n /= 10;
      if (cnt[k] == 0) ret++;
      cnt[k]++;
    }
    return ret;
  }
  int mind(int n) {
    std::vector<int> cnt(10, 0);
    while (n > 0) {
      cnt[n % 10] = 1;
      n /= 10;
    }
    for (int i = 0; i < 10; i++)
      if (cnt[i] == 1) return i;
    return -1;
  }
  int get(int n, int k) {
    while (1) {
      if (test(n) <= k) return n;
      if (n % 10 == 0) {
        int res = get(n / 10, k);
        return 10 * res + mind(res);
      }
      n++;
    }
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
      int n, k;
      scanf("%d %d", &n, &k);
      printf("%d\n", get(n, k));
    }
  }
  