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 int64_t INF = 1e18, MOD = 998244353;
void add(int64_t &a, int64_t b) { a = (a + b) % MOD; }
int64_t pp(int64_t b, int64_t p) {
  if (p == 0) return 1;
  int64_t v = pp(b, p / 2);
  return (v * v) % MOD * (p % 2 ? b : 1) % MOD;
}
int64_t n;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  vector<int64_t> fact(n + 1, 1), invf(n + 1);
  for (int64_t i = 2; i < n + 1; i++) fact[i] = fact[i - 1] * i % MOD;
  for (int64_t i = 0; i < n + 1; i++) invf[i] = pp(fact[i], MOD - 2);
  vector<vector<int64_t> > c(n + 1, vector<int64_t>(n + 1, 0));
  c[1][0] = 1;
  for (int64_t i = 2; i < n + 1; i++) {
    for (int64_t j = 0; j < i; j++) {
      c[i][j] = (j + 1) * c[i - 1][j] % MOD;
      if (j) add(c[i][j], (i - j) * c[i - 1][j - 1]);
    }
  }
  for (int64_t i = 1; i < n + 1; i++) {
    int64_t ans = 0;
    for (int64_t j = i; j < n + 1; j++)
      add(ans, c[j][i - 1] * fact[n] % MOD * invf[j]);
    cout << ans << " ";
  }
  cout << endl;
  return 0;
}