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 n, c[5005][5005], p[5005][5005], fac[5005];
int Mod(int x) { return x < mod ? x : x - mod; }
int main() {
  cin >> n, fac[0] = 1;
  for (int i = 0; i <= n; ++i) p[i][0] = c[i][0] = 1;
  for (int i = 1; i <= n; ++i) {
    fac[i] = 1ll * fac[i - 1] * i % mod;
    for (long long j = 1; j <= i; ++j) {
      c[i][j] = Mod(c[i - 1][j] + c[i - 1][j - 1]);
      p[i][j] = ((j + 1) * p[i - 1][j] + (i - j) * p[i - 1][j - 1]) % mod;
    }
  }
  for (int i = 0; i < n; ++i) {
    int ans = 0;
    for (int j = max(i, 1); j <= n; ++j)
      ans = (ans + 1ll * p[j][i] * c[n][j] % mod * fac[n - j]) % mod;
    cout << ans << " ";
  }
  return 0;
}