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 = 5001, mod = 998244353;
long long n, dp[N][N];
long long inv(long long a) {
  int k = mod - 2;
  long long b = 1;
  while (k) {
    if (k % 2) b = b * a % mod;
    a = a * a % mod;
    k /= 2;
  }
  return b;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  dp[0][0] = 1;
  for (int i = 2; i <= n; i++) dp[0][0] = dp[0][0] * i % mod;
  for (int i = 1; i <= n; i++) {
    long long invi = inv(i);
    for (int j = 1; j <= n; j++) {
      dp[i][j] = ((i - j + 1) * dp[i - 1][j - 1] + j * dp[i - 1][j]) % mod *
                 invi % mod;
    }
  }
  for (int j = 1; j <= n; j++) {
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      ans += dp[i][j];
      if (ans >= mod) ans -= mod;
    }
    cout << ans << " ";
  }
  cout << "\n";
  return 0;
}