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_64 mrand(chrono::steady_clock::now().time_since_epoch().count());
long long t[5005][5005];
const int mod = 998244353;
long long ans[5005];
long long fpow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
int main() {
  t[1][1] = 1;
  for (int n = 2; n <= 5000; ++n) {
    for (int k = 1; k <= n; ++k) {
      t[n][k] = (k * t[n - 1][k] + (n - k + 1) * t[n - 1][k - 1]) % mod;
    }
  }
  long long f = 1;
  int N;
  cin >> N;
  for (int i = 1; i <= N; ++i) f = f * i % mod;
  for (int i = 1; i <= N; ++i) ans[i] = f;
  for (int i = 2; i <= N; ++i) {
    long long tmp = ans[i];
    ans[i] = 0;
    long long tot = 0;
    for (int j = 1; j <= i; ++j) tot += t[i][j];
    tot %= mod;
    tmp = tmp * fpow(tot, mod - 2) % mod;
    for (int j = 1; j <= i; ++j) ans[j] = (ans[j] + tmp * t[i][j] % mod) % mod;
  }
  for (int i = 1; i <= N; ++i) cout << ans[i] << " ";
  cout << '\n';
}