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>
  #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
  using namespace std;
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  void ioi(string name) {
    freopen((name + ".in").c_str(), "r", stdin);
    freopen((name + ".out").c_str(), "w", stdout);
  }
  long long n, w, k, dp[20][2][1 << 10][2], ans[100005];
  string s;
  long long f(long long a = n - 1, long long b = 1, long long c = (1LL << 10) - 1,
              long long no = 1) {
    if (__builtin_popcountll(c) < 10 - k) return 0;
    if (a < 0) return 1;
    long long &d = dp[a][b][c][no];
    if (!b && ~d) return d;
    long long e = s[a] - '0', g = 0;
    for (long long i = 0; i <= (b ? e : 9); i++) {
      if (!(no && !i))
        g += f(a - 1, b && (i == e), (c & ~(1LL << i)), no && !i);
      else
        g += f(a - 1, b && (i == e), c, no && !i);
    }
    return b ? g : d = g;
  }
  long long calc(long long x) {
    s = to_string(x);
    n = (int)s.size();
    reverse(s.begin(), s.end());
    return f();
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    cout.precision(30);
    cerr.precision(10);
    ;
    vector<pair<long long, long long>> q[11];
    long long T;
    cin >> T;
    for (int i = 1; i <= T; i++) {
      cin >> w >> k;
      q[k].push_back(make_pair(w, i));
    }
    for (int K = 1; K <= 10; K++) {
      memset(dp, -1, sizeof dp);
      k = K;
      for (auto x : q[K]) {
        w = x.first;
        long long l = w, r = (1LL << 42);
        w = calc(w - 1);
        while (l != r) {
          long long tm = (l + r) / 2;
          if (w < calc(tm))
            r = tm;
          else
            l = tm + 1;
        }
        ans[x.second] = l;
      }
    }
    for (int i = 1; i <= T; i++) cout << ans[i] << "\n";
    return 0;
  }
  