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>
int quick_power(int a, int b, int Mod) {
  int ans = 1;
  while (b) {
    if (b & 1) {
      ans = 1ll * ans * a % Mod;
    }
    b >>= 1;
    a = 1ll * a * a % Mod;
  }
  return ans;
}
const int Maxn = 10000;
const int Mod = 998244353;
int n;
int frac[Maxn + 5], inv_f[Maxn + 5];
int inv[Maxn + 5];
void init() {
  frac[0] = 1;
  for (int i = 1; i <= Maxn; i++) {
    frac[i] = 1ll * frac[i - 1] * i % Mod;
  }
  inv_f[Maxn] = quick_power(frac[Maxn], Mod - 2, Mod);
  for (int i = Maxn - 1; i >= 0; i--) {
    inv_f[i] = 1ll * inv_f[i + 1] * (i + 1) % Mod;
  }
  for (int i = 1; i <= Maxn; i++) {
    inv[i] = 1ll * inv_f[i] * frac[i - 1] % Mod;
  }
}
int f[Maxn + 5];
int ans[Maxn + 5];
int main() {
  init();
  scanf("%d", &n);
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    if (i != 1) {
      int now = (i >> 1);
      for (int j = now; j >= 0; j--) {
        f[j] = (1ll * (j + 1) * f[j] + 1ll * (i - j) * f[j - 1]) % Mod;
      }
      if (!(i & 1)) {
        f[now] = f[now - 1];
      }
    }
    for (int j = (i >> 1) - 1; j >= 0; j--) {
      int tmp = 1ll * f[j] * inv_f[i] % Mod;
      ans[j] = (ans[j] + tmp) % Mod;
      ans[i - j - 1] = (ans[i - j - 1] + tmp) % Mod;
    }
    if (i & 1) {
      ans[i >> 1] = (ans[i >> 1] + 1ll * f[i >> 1] * inv_f[i]) % Mod;
    }
  }
  for (int i = 0; i < n; i++) {
    int tmp = 1ll * ans[i] * frac[n] % Mod;
    printf("%d ", tmp);
  }
  puts("");
  return 0;
}