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 Maxx = 5005, Mod = 998244353;
int n, er[Maxx], Inv[Maxx], Jc[Maxx], f[Maxx];
inline long long Qpow(long long x, int p = Mod - 2) {
  static long long res;
  for (res = 1; p; p >>= 1, x = x * x % Mod)
    if (p & 1) res = res * x % Mod;
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = Jc[0] = 1; i <= n; ++i) Jc[i] = (long long)i * Jc[i - 1] % Mod;
  Inv[n] = Qpow(Jc[n], Mod - 2);
  for (int i = n; i; --i) Inv[i - 1] = (long long)i * Inv[i] % Mod;
  for (int i = er[0] = 1; i <= n; ++i) {
    for (int j = i; j; --j)
      er[j] =
          ((long long)(j + 1) * er[j] + (long long)(i - j) * er[j - 1]) % Mod;
    for (int j = 1; j <= i; ++j)
      f[j] = (f[j] + (long long)er[j - 1] % Mod * Inv[i]) % Mod;
  }
  for (int i = 1; i <= n; ++i) cout << (long long)f[i] * Jc[n] % Mod << " ";
  puts("");
  return 0;
}