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 Mod = 998244353;
int f[5010][5010];
int C[5010][5010];
int fac[5010];
int main() {
  int n;
  scanf("%d", &n);
  f[0][0] = 1;
  C[0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= i; j++) {
      f[i + 1][j] = (f[i + 1][j] + 1ll * f[i][j] * j) % Mod;
      f[i + 1][j + 1] = (f[i + 1][j + 1] + 1ll * f[i][j] * (i - j + 1)) % Mod;
      C[i + 1][j] = (C[i + 1][j] + C[i][j]) % Mod;
      C[i + 1][j + 1] = (C[i + 1][j + 1] + C[i][j]) % Mod;
    }
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
  for (int i = 1; i <= n; i++) {
    int ans = 0;
    for (int j = i; j <= n; j++)
      ans = (ans + 1ll * C[n][j] * f[j][i] % Mod * fac[n - j]) % Mod;
    printf("%d ", ans);
  }
  printf("\n");
  return 0;
}