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>
const int mo = 998244353, N = 5005;
int n, f[N][N], i, j, ans[N];
int vf[N], fac[N], inv[N];
inline int C(int n, int m) {
  return 1ll * vf[m] * vf[n - m] % mo * fac[n] % mo;
}
int main() {
  for (int i = *vf = *fac = 1; i < N; ++i)
    inv[i] = i == 1 ? 1 : 1ll * (mo - mo / i) * inv[mo % i] % mo,
    vf[i] = 1ll * vf[i - 1] * inv[i] % mo, fac[i] = 1ll * fac[i - 1] * i % mo;
  scanf("%d", &n);
  f[0][0] = 1;
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= i; ++j) {
      f[i][j] =
          (1ll * j * f[i - 1][j] + 1ll * (i - j + 1) * f[i - 1][j - 1]) % mo;
      ans[j] = (ans[j] + 1ll * f[i][j] * C(n, i) % mo * fac[n - i]) % mo;
    }
  for (i = 1; i <= n; ++i) printf("%d%c", ans[i], i == n ? '\n' : ' ');
}