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 int maxn = 500005;
  const long long Mod = 998244353;
  int dp[1030][12];
  int a[20];
  int n;
  int pw[maxn];
  int k;
  long long dfs(int pos, long long pre, int st, int now, int fl) {
    if (pos > n) return pre;
    if (fl == 1) {
      if (now < k) {
        if (st & 1) return dfs(pos + 1, 10ll * pre, st, now, fl);
        return dfs(pos + 1, 10ll * pre, st | 1, now + 1, fl);
      }
      for (int i = 0; i <= 9; i++) {
        if (st & pw[i]) return dfs(pos + 1, 10ll * pre + i, st, now, fl);
      }
      return 1e18;
    }
    long long mn = 1e18;
    if (st & pw[a[pos]])
      mn = min(mn, dfs(pos + 1, 10ll * pre + a[pos], st, now, fl));
    else {
      if (now < k) {
        mn = min(mn,
                 dfs(pos + 1, 10ll * pre + a[pos], st | pw[a[pos]], now + 1, fl));
      }
    }
    for (int i = a[pos] + 1; i <= 9; i++) {
      if (st & pw[i]) {
        mn = min(mn, dfs(pos + 1, 10ll * pre + i, st, now, 1));
        break;
      } else {
        if (now < k) {
          mn = min(mn, dfs(pos + 1, 10ll * pre + i, st | pw[i], now + 1, 1));
        }
      }
    }
    return mn;
  }
  int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    ;
    pw[0] = 1;
    for (int i = 1; i <= 9; i++) pw[i] = 2 * pw[i - 1];
    int t;
    cin >> t;
    while (t--) {
      long long x;
      cin >> x >> k;
      n = 0;
      while (x) {
        a[++n] = x % 10;
        x /= 10;
      }
      reverse(a + 1, a + n + 1);
      cout << dfs(1, 0, 0, 0, 0) << endl;
    }
    return 0;
  }
  