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 int N = 5003;
const int mod = 998244353;
int n;
int fac[N], C[N][N], E[N][N];
int ans[N];
void init() {
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  for (int i = 0; i <= n; i++) C[i][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
  E[1][1] = 1;
  for (int i = 2; i <= n; i++)
    for (int j = 1; j <= i; j++)
      E[i][j] = (1ll * E[i - 1][j - 1] * (i - j + 1) % mod +
                 1ll * E[i - 1][j] * j % mod) %
                mod;
}
int main() {
  scanf("%d", &n);
  init();
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++)
      ans[i] =
          (ans[i] + 1ll * C[n][j] % mod * fac[n - j] % mod * E[j][i] % mod) %
          mod;
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}