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 N = 11;
  const int M = (1 << 10) + 1;
  int n, k;
  bool dp[N][M][2];
  bool vis[N][M][2];
  int jabo[N][M][2];
  vector<int> digits;
  void fun(int n) {
    while (n != 0) {
      digits.push_back(n % 10);
      n /= 10;
    }
    reverse(digits.begin(), digits.end());
  }
  bool solve(int idx, int mask, bool done) {
    if (__builtin_popcount(mask) > k) return 0;
    if (idx == (int)digits.size()) return 1;
    bool &ret = dp[idx][mask][done];
    if (vis[idx][mask][done]) {
      return ret;
    }
    bool ans = false;
    if (!done) {
      for (int i = digits[idx]; i <= 9; i++) {
        bool eta = 0;
        if (i != digits[idx]) eta = 1;
        if (solve(idx + 1, mask | (1 << i), eta)) {
          jabo[idx][mask][done] = i;
          ans = 1;
          break;
        }
      }
    } else {
      for (int i = 0; i <= 9; i++) {
        if (solve(idx + 1, mask | (1 << i), 1)) {
          jabo[idx][mask][done] = i;
          ans = 1;
          break;
        }
      }
    }
    return ret = ans;
  }
  void print(int idx, int mask, bool done) {
    if (idx == digits.size()) return;
    int i = jabo[idx][mask][done];
    cout << i;
    mask |= (1 << i);
    if (!done && i > digits[idx]) done = 1;
    print(idx + 1, mask, done);
  }
  void TEST_CASES() {
    memset(vis, 0, sizeof vis);
    digits.clear();
    cin >> n >> k;
    fun(n);
    bool x = solve(0, 0, 0);
    print(0, 0, 0);
    cout << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;
    cin >> t;
    while (t--) {
      TEST_CASES();
    }
    return 0;
  }
  