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 long long MOD = 998244353;
int N;
long long f[5001][5001], O[5001], inv[5001], fi[5001], fc[5001];
int main() {
  scanf("%d", &N);
  f[1][0] = 1;
  for (int i = 1; i < N; i++)
    for (int j = 0; j <= i - 1; j++) {
      f[i + 1][j + 1] = (f[i + 1][j + 1] + f[i][j] * (i - j)) % MOD;
      f[i + 1][j] = (f[i + 1][j] + f[i][j] * (j + 1)) % MOD;
    }
  inv[1] = 1;
  for (int i = 2; i <= N; i++) inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
  fc[0] = fi[0] = 1;
  for (int i = 1; i <= N; i++) {
    fc[i] = fc[i - 1] * i % MOD;
    fi[i] = fi[i - 1] * inv[i] % MOD;
  }
  for (int i = 1; i <= N; i++) {
    for (int j = i; j <= N; j++)
      O[i] = (O[i] + f[j][i - 1] * fc[N] % MOD * fi[j]) % MOD;
    printf("%lld%c", O[i], " \n"[i == N]);
  }
  return 0;
}