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 MX = 5005, mod = 998244353;
int n;
long long fac[MX], inv[MX], mem[MX][MX];
long long pot(long long b, int p) {
  long long res = 1;
  while (p) {
    if (p & 1) (res *= b) %= mod;
    (b *= b) %= mod;
    p /= 2;
  }
  return res;
}
long long dp(int n, int k) {
  if (k == 1) return 1;
  if (n == 0) return 0;
  long long &res = mem[n][k];
  if (res != -1) return res;
  return res = (k * dp(n - 1, k) + (n - k + 1) * dp(n - 1, k - 1)) % mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  memset(mem, -1, sizeof(mem));
  fac[0] = inv[0] = 1;
  for (int i = 1; i < MX; i++) {
    fac[i] = i * fac[i - 1] % mod;
    inv[i] = pot(fac[i], mod - 2);
  }
  cin >> n;
  for (int i = 1; i <= n; i++) {
    long long res = 0;
    for (int j = i; j <= n; j++)
      (res += dp(j, i) * fac[n] % mod * inv[j]) %= mod;
    cout << ((((res) % (mod)) + (mod)) % (mod)) << " ";
  }
  cout << '\n';
  return 0;
}