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 Inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const int jt = 998244353;
int inline ksmii(int a, int b) {
  if (!b) return 1;
  int x = ksmii(a, b >> 1);
  x = 1ll * x * x % jt;
  if (b & 1) x = 1ll * x * a % jt;
  return x;
}
int n;
int jc[100005], fjc[100005];
int ora[5005][5005];
int main() {
  scanf("%d", &n);
  jc[0] = 1;
  for (int i = 1; i <= (n); ++i) jc[i] = 1ll * i * jc[i - 1] % jt;
  fjc[n] = ksmii(jc[n], jt - 2);
  for (int i = (n); (-1) > 0 ? i <= (1) : i >= (1); i += (-1))
    fjc[i - 1] = 1ll * i * fjc[i] % jt;
  for (int i = 1; i <= (n + 1); ++i) ora[i][0] = 1;
  for (int i = 1; i <= (n); ++i)
    for (int j = 1; j <= (i); ++j)
      ora[i][j] =
          (1ll * (i - j) * ora[i - 1][j - 1] + 1ll * (j + 1) * ora[i - 1][j]) %
          jt;
  for (int i = 0; i < (n); ++i) {
    int ans = 0;
    for (int j = (max(i, 1)); j <= (n); ++j)
      ans = (ans + 1ll * jc[n] * fjc[j] % jt * ora[j][i]) % jt;
    printf("%d ", ans);
  }
  return 0;
}