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 a[20];
  int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      vector<int> v;
      int y = n;
      while (y) {
        v.push_back(y % 10);
        y /= 10;
      }
      reverse(v.begin(), v.end());
      int sz = v.size();
      if (k == 1) {
        bool les = 0, bg = 0;
        for (int i = 1; i < sz; i++) {
          if (v[i] < v[0])
            les = 1;
          else if (v[i] > v[0]) {
            bg = 1;
            break;
          }
        }
        if (!les && bg)
          for (int i = 0; i < sz; i++) a[i] = v[0] + 1;
        else
          for (int i = 0; i < sz; i++) a[i] = v[0];
        for (int i = 0; i < sz; i++) cout << a[i];
        cout << '\n';
        continue;
      }
      int cnt[11] = {0};
      int cc = 0;
      for (auto i : v)
        if (!cnt[i]) cc++, cnt[i]++;
      if (cc <= k) {
        cout << n << '\n';
        continue;
      }
      for (int i = 0; i < 10; i++) cnt[i] = 0;
      cc = 0;
      int mn = 10, mx = -1;
      for (int i = 0; i < sz; i++) {
        int x = v[i];
        if (!cnt[x]) {
          cc++;
        }
        if (cc > k) {
          if (x > mx) {
            int sum = 0;
            for (int j = i - 1; j >= 0; j--) {
              if (v[j] != mx) break;
              sum++;
            }
            if (sum == cnt[mx]) {
              for (int j = i - 1; j >= 0; j--) {
                if (v[j] == mx)
                  a[j] = mn;
                else {
                  a[j + 1] = mx + 1;
                  break;
                }
              }
              for (int j = i; j < sz; j++) a[j] = mn;
            } else {
              for (int j = i - 1; j >= 0; j--) {
                if (v[j] != mx) {
                  int y, nx;
                  if (cnt[v[j]] == 1) {
                    y = v[j] + 1, nx;
                    if (cnt[y]) {
                      nx = 0;
                    } else {
                      if (v[j] == mn)
                        nx = y;
                      else
                        nx = mn;
                    }
                  } else {
                    for (y = v[j] + 1; y < 10; y++) {
                      if (cnt[y]) break;
                    }
                    nx = mn;
                  }
                  a[j] = y;
                  for (int q = j + 1; q < sz; q++) {
                    a[q] = nx;
                  }
                  break;
                }
              }
            }
          } else {
            for (int j = x + 1; j < 10; j++) {
              if (cnt[j]) {
                a[i] = j;
                break;
              }
            }
            for (int j = i + 1; j < sz; j++) a[j] = mn;
          }
          break;
        }
        a[i] = x;
        cnt[x]++;
        mn = min(mn, x), mx = max(mx, x);
      }
      for (int i = 0; i < sz; i++) cout << a[i];
      cout << '\n';
    }
    return 0;
  }
  