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;
  int k, cnt[1024], len, n, ans;
  char st[15];
  void dfs(bool limit, int pos, int Bit, int x) {
    if (ans != INT_MAX) return;
    if (pos == len + 1) {
      if (x >= n) ans = min(ans, x);
      return;
    }
    int min = INT_MAX;
    for (int i = (limit ? (st[pos] - '0') : 0); i <= 9; ++i) {
      if (cnt[Bit | (1 << i)] <= k)
        dfs(limit & (i == (st[pos] - '0')), pos + 1, Bit | (1 << i), x * 10 + i);
    }
  }
  int main() {
    int tmp;
    scanf("%d", &tmp);
    cnt[0] = 0;
    for (int i = 1; i < 1024; ++i) cnt[i] = cnt[i - (i & (-i))] + 1;
    while (tmp--) {
      scanf("%s", st + 1);
      len = strlen(st + 1);
      n = 0;
      for (int i = 1; i <= len; ++i) n = n * 10 + (st[i] - '0');
      scanf("%d", &k);
      ans = INT_MAX;
      dfs(1, 1, 0, 0);
      printf("%d\n", ans);
    }
    return 0;
  }
  