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>
const int Mod = 998244353;
const int MN = 5005;
int N;
int E[MN][MN];
int main() {
  scanf("%d", &N);
  E[1][1] = 1;
  for (int i = 2; i <= N; ++i)
    for (int j = 1; j <= i; ++j)
      E[i][j] = ((long long)(i - j + 1) * E[i - 1][j - 1] +
                 (long long)j * E[i - 1][j]) %
                Mod;
  for (int i = 1; i <= N; ++i) {
    int Sum = 0, Coef = 1;
    for (int j = N; j >= 1; Coef = (long long)Coef * j % Mod, --j)
      Sum = (Sum + (long long)Coef * E[j][i]) % Mod;
    printf("%d%c", Sum, " \n"[i == N]);
  }
  return 0;
}