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;
namespace FGF {
int n, m;
const int N = 5005, mo = 998244353;
int o[N][N], fac, inv[N], ans[N];
void work() {
  scanf("%d", &n);
  fac = o[0][0] = inv[0] = inv[1] = 1;
  for (int i = 2; i <= n; i++)
    inv[i] = 1ll * inv[mo % i] * (mo - mo / i) % mo, fac = 1ll * fac * i % mo;
  for (int i = 2; i <= n; i++) inv[i] = 1ll * inv[i] * inv[i - 1] % mo;
  for (int i = 1; i <= n; i++) {
    o[i][0] = 1;
    for (int j = 1; j <= i; j++)
      o[i][j] = (1ll * (i - j) * o[i - 1][j - 1] % mo +
                 1ll * (j + 1) * o[i - 1][j] % mo) %
                mo;
  }
  for (int j = 1; j <= n; j++)
    for (int i = 1; i <= n; i++)
      ans[j] = (ans[j] + 1ll * o[i][j - 1] * fac % mo * inv[i] % mo) % mo;
  for (int j = 1; j <= n; j++) printf("%d ", ans[j]);
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}