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 = 3e5 + 5;
  int ans, ans1, fix[100], n1, n, k, raodd, cnt, t, pw[20], bo, cur, cntt, ff,
      fi[20];
  vector<int> v, vv;
  signed main() {
    std::ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> t;
    pw[0] = 1;
    ans = 1e9;
    for (int i = 1; i <= 10; i++) {
      pw[i] = pw[i - 1] * 10;
    }
    while (t--) {
      cin >> n >> k;
      if (n == 1000000000) {
        if (k == 1) {
          cout << 1111111111 << endl;
        } else
          cout << 1000000000 << endl;
        continue;
      }
      n1 = n;
      for (int i = 0; i <= 9; i++) {
        fix[i] = 0;
      }
      raodd = 0;
      cntt = 0;
      ans = 1e9;
      while (n1) {
        v.push_back(n1 % 10);
        if (!fix[n1 % 10]) raodd++, fix[n1 % 10] = 1;
        n1 /= 10;
        cntt++;
      }
      reverse(v.begin(), v.end());
      if (raodd <= k) {
        cout << n << " ";
        continue;
      }
      for (int i = 1; i < (1 << 10); i++) {
        vv.clear();
        cnt = 0;
        ans1 = 0;
        for (int j = 0; j <= 9; j++) {
          if ((1 << j) & i) {
            vv.push_back(j);
            cnt++;
          }
        }
        for (int kk = 0; kk <= 9; kk++) fi[kk] = 0;
        for (int kk = 0; kk < vv.size(); kk++) {
          fi[vv[kk]] = 1;
        }
        if (cnt > k) continue;
        for (int idx = 0; idx < cntt; idx++) {
          bo = 0;
          ans1 = 0;
          ff = 0;
          for (int jj = 0; jj < idx; jj++) {
            if (fi[v[jj]])
              ans1 += pw[cntt - jj - 1] * v[jj];
            else {
              bo = 1;
              break;
            }
          }
          if (bo) break;
          ff = 0;
          for (int kk = 0; kk < vv.size(); kk++) {
            if (vv[kk] > v[idx]) {
              ff = 1;
              ans1 += pw[cntt - idx - 1] * vv[kk];
              break;
            }
          }
          if (!ff) continue;
          for (int jj = idx + 1; jj < cntt; jj++) {
            ans1 += pw[cntt - jj - 1] * vv[0];
          }
          if (ans1 < 0) continue;
          ans = min(ans, ans1);
        }
      }
      cout << ans << " ";
    }
  }
  