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;
int n;
long long d[5020][5020], res[5020], inv[5020], fac[5020], ifac[5020],
    MOD = 998244353;
void init_math(int N) {
  inv[1] = 1LL;
  for (int i = 2; i <= N; i++)
    inv[i] = (MOD - ((MOD / i) * inv[MOD % i]) % MOD);
  fac[0] = 1LL;
  ifac[0] = 1LL;
  for (int i = 1; i <= N; i++) {
    fac[i] = (fac[i - 1] * (long long)(i)) % MOD;
    ifac[i] = (ifac[i - 1] * inv[i]) % MOD;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  init_math(n);
  d[0][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++)
      d[i][j] = (d[i - 1][j - 1] * (long long)(i - j + 1) +
                 d[i - 1][j] * (long long)(j)) %
                MOD;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) res[i] += (d[j][i] * ifac[j]) % MOD;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) res[i] = (res[i] * j) % MOD;
  for (int i = 1; i <= n; i++) cout << res[i] << ' ';
  cout << endl;
  return 0;
}