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 Mod = 998244353;
int fpow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1ll * res * x % Mod;
    x = 1ll * x * x % Mod, y >>= 1;
  }
  return res;
}
const int MAX_N = 5005;
int N, E[MAX_N][MAX_N];
int fac[MAX_N], ifc[MAX_N];
int C(int n, int m) {
  if (n < m || n < 0 || m < 0)
    return 0;
  else
    return 1ll * fac[n] * ifc[m] % Mod * ifc[n - m] % Mod;
}
int main() {
  cin >> N;
  for (int i = E[0][0] = 1; i <= N; i++)
    for (int j = E[i][0] = 1; j < i; j++)
      E[i][j] =
          (1ll * E[i - 1][j] * (j + 1) + 1ll * (i - j) * E[i - 1][j - 1]) % Mod;
  for (int i = fac[0] = 1; i <= N; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
  ifc[N] = fpow(fac[N], Mod - 2);
  for (int i = N - 1; ~i; i--) ifc[i] = 1ll * ifc[i + 1] * (i + 1) % Mod;
  for (int k = 1; k <= N; k++) {
    int ans = 0;
    for (int i = k; i <= N; i++)
      ans = (ans + 1ll * E[i][k - 1] * C(N, i) % Mod * fac[N - i]) % Mod;
    printf("%d ", ans);
  }
  putchar('\n');
  return 0;
}