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 MAXN = 5100;
const long long MOD = 998244353;
int N;
long long ccnt[MAXN][MAXN];
long long cfac[MAXN];
long long ifac[MAXN];
long long cpow(long long b, int e) {
  if (e == 0) return 1;
  if (e == 1) return (b % MOD);
  long long vv = cpow(b, e / 2);
  vv = (vv * vv) % MOD;
  if (e & 1) vv = (vv * b) % MOD;
  return vv;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N;
  ccnt[1][1] = 1;
  cfac[0] = 1;
  for (int i = 1; i < MAXN; i++) cfac[i] = (i * cfac[i - 1]) % MOD;
  ifac[MAXN - 1] = cpow(cfac[MAXN - 1], MOD - 2);
  for (int i = MAXN - 2; i >= 0; i--) ifac[i] = (ifac[i + 1] * (i + 1)) % MOD;
  for (int j = 1; j < N; j++) {
    for (int i = 1; i <= j; i++) {
      ccnt[i][j + 1] += i * ccnt[i][j];
      ccnt[i][j + 1] %= MOD;
      ccnt[i + 1][j + 1] += (j - i + 1) * ccnt[i][j];
      ccnt[i + 1][j + 1] %= MOD;
    }
  }
  for (int i = 1; i <= N; i++) {
    long long res = 0;
    for (int j = 1; j <= N; j++) {
      long long gg = cfac[N] * ifac[j];
      res = (res + ccnt[i][j] * (gg % MOD)) % MOD;
    }
    if (i > 1) cout << " ";
    cout << res % MOD;
  }
  cout << "\n";
}