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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int maxn = 3e6;
const long long mod = 998244353;
const long double PI = acos((long double)-1);
long long pw(long long a, long long b, long long md = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (a * res) % md;
    }
    a = (a * a) % md;
    b >>= 1;
  }
  return (res);
}
long long n, dp[5100][5100], ans;
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    dp[i][0] = 1;
    for (int j = 1; j < i; j++) {
      dp[i][j] = (dp[i - 1][j - 1] * (i - j) + dp[i - 1][j] * (j + 1)) % mod;
    }
  }
  for (int i = 1; i <= n; i++) {
    ans = 1;
    for (int j = i + 1; j <= n; j++) ans = (ans * j + dp[j][i - 1]) % mod;
    cout << ans << ' ';
  }
  return (0);
}