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 long long p = 998244353;
long long n, ans, jc[100010], ny[100010], f[5010][5010];
long long ksm(long long x, long long y) {
  long long xlh = 1;
  while (y) {
    if (y & 1) xlh = xlh * x % p;
    x = x * x % p;
    y /= 2;
  }
  return xlh;
}
long long C(long long x, long long y) {
  return jc[x] * ny[y] % p * ny[x - y] % p;
}
int main() {
  long long i, j;
  scanf("%lld", &n);
  jc[0] = ny[0] = 1;
  for (i = 1; i <= n; i++) jc[i] = jc[i - 1] * i % p, ny[i] = ksm(jc[i], p - 2);
  f[1][1] = 1;
  for (i = 2; i <= n; i++)
    for (j = 1; j <= i; j++)
      f[i][j] = ((i - j + 1) * f[i - 1][j - 1] + j * f[i - 1][j]) % p;
  for (i = 1; i <= n; i++) {
    ans = 0;
    for (j = i; j <= n; j++) {
      ans = (ans + f[j][i] * C(n, j) % p * jc[n - j]) % p;
    }
    printf("%lld ", ans);
  }
}