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>
int n, f[5001][5001], ans[5001], inv[5001];
signed main(void) {
  int i, j, jc;
  scanf("%d", &n);
  for (*inv = inv[1] = 1, i = 2; i <= n; ++i)
    inv[i] = 1ll * (998244353 - 998244353 / i) * inv[998244353 % i] % 998244353;
  for (jc = i = 1; i <= n; ++i)
    jc = 1ll * jc * i % 998244353,
    inv[i] = 1ll * inv[i - 1] * inv[i] % 998244353;
  for (f[0][0] = i = 1; i <= n; ++i)
    for (f[i][0] = j = 1; j <= i; ++j)
      f[i][j] = (f[i][j] + 1ll * (i - j) * f[i - 1][j - 1] +
                 1ll * (j + 1) * f[i - 1][j]) %
                998244353;
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j)
      ans[j] = (ans[j] + 1ll * inv[i] * f[i][j - 1]) % 998244353;
  for (i = 1; i <= n; ++i) printf("%lld ", 1ll * ans[i] * jc % 998244353);
  return 0;
}