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 = 5005, mod = 998244353;
int n, C[N][N], dp[N][N], ans[N], fac[N];
void add(int x, int c) {
  dp[x + 1][c] = (dp[x + 1][c] + 1ll * dp[x][c] * (c + 1) % mod) % mod;
  dp[x + 1][c + 1] = (dp[x + 1][c + 1] + 1ll * dp[x][c] * (x - c) % mod) % mod;
}
int main() {
  cin >> n;
  fac[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  for (int i = 0; i < N; i++) {
    C[0][i] = 1;
    for (int j = 1; j < i + 1; j++)
      C[j][i] = (C[j][i - 1] + C[j - 1][i - 1]) % mod;
  }
  dp[1][0] = 1;
  for (int i = 1; i < n + 1; i++)
    for (int j = 0; j < i; j++) add(i, j);
  for (int i = 1; i < n + 1; i++)
    for (int j = 1; j < n + 1; j++)
      ans[j] =
          (ans[j] + 1ll * dp[i][j - 1] * C[i][n] % mod * fac[n - i] % mod) %
          mod;
  for (int i = 1; i < n + 1; i++) cout << ans[i] << " ";
}