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 ff[5005][5005], dp[5005][5005];
int main() {
  for (int i = 1; i < 5005; i++) {
    int nw = 1;
    for (long long j = i; j < 5005; j++) {
      nw = (nw * 1ll * j) % mod;
      if (nw < 0) nw += mod;
      ff[i][(int)j] = nw;
    }
  }
  dp[1][0] = 1;
  for (int i = 2; i < 5005; i++)
    for (int j = 0; j < i; j++) {
      dp[i][j] = ((j + 1) * 1ll * dp[i - 1][j]) % mod;
      if (dp[i][j] < 0) dp[i][j] += mod;
      if (j > 0)
        dp[i][j] = (dp[i][j] + ((i - j) * 1ll * dp[i - 1][j - 1]) % mod) % mod;
      if (dp[i][j] < 0) dp[i][j] += mod;
    }
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    long long ans = 0ll;
    for (int j = 1; j <= n; j++) {
      ans = (ans +
             (dp[j][i - 1] * 1ll * (j + 1 <= n ? ff[j + 1][n] : 1ll)) % mod) %
            mod;
      if (ans < 0) ans += mod;
    }
    cout << ans << " ";
  }
  cout << endl;
  return 0;
}