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 i64 = long long;
  vector<int> lim;
  i64 dp[13][1 << 10][2][11];
  i64 solve(int p, int mask, int k, int s, int tight) {
    if (__builtin_popcount(mask) > k) return 0;
    if (p < 0) return 1;
    i64& res = dp[p][mask][s][k];
    if (tight && ~res) return res;
    i64 cnt = 0;
    for (int i = 0; i <= (tight ? 9 : lim[p]); i++) {
      if (!s && i == 0)
        cnt += solve(p - 1, 0, k, 0, 1);
      else
        cnt += solve(p - 1, mask | (1 << i), k, 1, tight | (i < lim[p]));
    }
    return (tight ? res = cnt : cnt);
  }
  i64 getSolve(i64 n, int k) {
    lim.clear();
    while (n) {
      lim.emplace_back(n % 10);
      n /= 10;
    }
    return solve((int)lim.size() - 1, 0, k, 0, 0);
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int tt;
    cin >> tt;
    memset(dp, -1, sizeof(dp));
    while (tt--) {
      int n, k;
      cin >> n >> k;
      i64 def = getSolve(n - 1, k);
      i64 lo = n, hi = 1e12;
      while (lo < hi) {
        const i64 mid = lo + (hi - lo) / 2;
        if (getSolve(mid, k) > def)
          hi = mid;
        else
          lo = mid + 1;
      }
      cout << hi << '\n';
    }
    return 0;
  }
  