Problem: Note that the only differences between easy and hard versions are the constraints on n and the time limit. You can make hacks only if all versions are solved.

Slime is interested in sequences. He defined good positive integer sequences p of length n as follows:

  * For each k>1 that presents in p, there should be at least one pair of indices i,j, such that 1 ≤ i < j ≤ n, p_i = k - 1 and p_j = k.



For the given integer n, the set of all good sequences of length n is s_n. For the fixed integer k and the sequence p, let f_p(k) be the number of times that k appears in p. For each k from 1 to n, Slime wants to know the following value:

$$$\left(∑_{p∈ s_n} f_p(k)\right)\ mod\ 998 244 353$$$

Input

The first line contains one integer n\ (1≤ n≤ 5000).

Output

Print n integers, the i-th of them should be equal to \left(∑_{p∈ s_n} f_p(i)\right)\ mod\ 998 244 353.

Examples

Input


2


Output


3 1 


Input


3


Output


10 7 1 


Input


1


Output


1 

Note

In the first example, s=\{[1,1],[1,2]\}.

In the second example, s=\{[1,1,1],[1,1,2],[1,2,1],[1,2,2],[2,1,2],[1,2,3]\}.

In the third example, s=\{[1]\}.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long P = 998244353;
long long t[5100][5100];
long long a[5100];
signed main() {
  t[1][1] = 1;
  for (long long n = 2; n <= 5000; ++n) {
    for (long long k = 1; k <= n; ++k) {
      t[n][k] = (t[n - 1][k] * k + t[n - 1][k - 1] * (n - k + 1)) % P;
    }
  }
  long long n;
  cin >> n;
  a[1] = 1;
  for (long long i = 2; i <= n; ++i) {
    for (long long j = 1; j <= i; ++j) {
      a[j] = (a[j] * i) % P;
      a[j] += t[i][j];
      a[j] %= P;
    }
  }
  for (long long i = 1; i <= n; ++i) {
    cout << a[i] << " ";
  }
  cout << "\n";
  return 0;
}