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 inf = 1000000007;
const long long INF = 1e18;
int mod = 998244353;
template <typename T1, typename T2>
inline bool chmin(T1& a, T2 b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool chmax(T1& a, T2 b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
template <typename T, typename U>
T pow_(T a, U b) {
  return b ? pow_(a * a, b / 2) * (b % 2 ? a : 1) : 1;
}
long long modpow(long long a, long long b, long long _mod) {
  return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1;
}
template <class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << p.first << " " << p.second;
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& vec) {
  for (int i = 0; i < ((int)vec.size()); ++i) {
    if (i) os << " ";
    os << vec[i];
  }
  return os;
}
template <typename T>
inline istream& operator>>(istream& is, vector<T>& v) {
  for (int j = 0; j < ((int)v.size()); ++j) is >> v[j];
  return is;
}
template <class T, class T2>
inline void add(T& a, T2 b) {
  a += b;
  if (a >= mod) a -= mod;
}
template <class T>
void operator+=(vector<T>& v, vector<T> v2) {
  for (int i = 0; i < ((int)v2.size()); ++i) v.emplace_back(v2[i]);
}
void solve();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(11);
  int T;
  T = 1;
  while (T--) solve();
}
void solve() {
  int n;
  cin >> n;
  vector<vector<long long> > dp(n + 1, vector<long long>(n + 1));
  dp[0][0] = 1;
  for (int i = 1; i < (n + 1); ++i)
    for (int j = 1; j < (i + 1); ++j)
      dp[i][j] = (dp[i - 1][j - 1] * (i - j + 1) + dp[i - 1][j] * j) % mod;
  vector<long long> f(n + 1);
  f[n] = 1;
  f[0] = 1;
  for (int i = n - 1; i >= 1; --i) f[i] = f[i + 1] * (i + 1) % mod;
  vector<long long> ans(n + 1);
  for (int i = 1; i < (n + 1); ++i) {
    for (int j = 1; j < (n + 1); ++j) add(ans[i], dp[j][i] * f[j] % mod);
  }
  for (int i = 1; i < (n + 1); ++i) cout << ans[i] << '\n';
}