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 = 5000 + 10, P = 998244353;
int n;
int dp[N][N], cnt[N], fac[N][N], fact[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n;
  fact[0] = 1;
  for (int i = 0; i <= n; i++) {
    dp[i][0] = fac[i][0] = 1;
    for (int j = 1; j < i; j++) {
      dp[i][j] = (1ll * dp[i - 1][j] * (j + 1) % P +
                  1ll * dp[i - 1][j - 1] * (i - j) % P) %
                 P;
    }
    for (int j = 1; j <= i; j++) {
      fac[i][j] = (fac[i - 1][j - 1] + fac[i - 1][j]) % P;
    }
  }
  for (int i = 1; i <= n; i++) fact[i] = 1ll * fact[i - 1] * i % P;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < i; j++) {
      (cnt[j + 1] += 1ll * fac[n][i] * dp[i][j] % P * fact[n - i] % P) %= P;
    }
  }
  for (int i = 1; i <= n; i++) cout << cnt[i] << " ";
  cout << endl;
  return 0;
}