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;
int n;
int fac[5005], C[5005][5005], E[5005][5005];
int main() {
  scanf("%d", &n);
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = (long long)fac[i - 1] * i % 998244353;
  for (int i = 0; i <= n; ++i) {
    C[i][0] = 1;
    for (int j = 1; j <= i; ++j)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 998244353;
  }
  E[0][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < i; ++j)
      E[i][j] = ((long long)(j + 1) * E[i - 1][j] +
                 (long long)(i - j) * E[i - 1][j - 1]) %
                998244353;
  for (int i = 1; i <= n; ++i) {
    long long ans = 0;
    for (int j = 1; j <= n; ++j)
      (ans += (long long)E[j][j - i] * C[n][j] % 998244353 * fac[n - j]) %=
          998244353;
    printf("%lld ", ans);
  }
  return 0;
}