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 int N = 5e3 + 10;
const int P = 998244353;
int s2[N][N];
int c[N][N];
int fact[N];
inline void add(int &x, int y) {
  x += y;
  if (x >= P) x -= P;
}
inline void init() {
  c[0][0] = s2[0][0] = fact[0] = 1;
  for (int i = 1; i <= N - 1; ++i)
    c[i][0] = 1, s2[i][0] = 0, fact[i] = 1LL * fact[i - 1] * i % P;
  for (int i = 1; i <= N - 1; ++i)
    for (int j = 1; j <= i; ++j) {
      add(c[i][j] = c[i - 1][j], c[i - 1][j - 1]);
      add(s2[i][j] = 1LL * j * s2[i - 1][j] % P, s2[i - 1][j - 1]);
    }
}
int n, ans[N], res[N];
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    for (int j = i - 1; j >= 0; --j) {
      add(res[j],
          1LL * c[n][i] * s2[i][i - j] % P * fact[i - j] % P * fact[n - i] % P);
    }
  }
  for (int j = n - 1; j >= 0; --j) {
    for (int w = n - 1; w >= j + 1; --w)
      add(res[j], P - 1LL * c[w][w - j] * res[w] % P);
    add(ans[j + 1], res[j]);
  }
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  return 0;
}