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>
const int mod = 998244353;
const int gmod = 3;
const int inf = 1039074182;
const double eps = 1e-9;
const long long llinf = 2LL * inf * inf;
template <typename T1, typename T2>
inline void chmin(T1 &x, T2 b) {
  if (b < x) x = b;
}
template <typename T1, typename T2>
inline void chmax(T1 &x, T2 b) {
  if (b > x) x = b;
}
inline void chadd(int &x, int b) {
  x += b - mod;
  x += (x >> 31 & mod);
}
template <typename T1, typename T2>
inline void chadd(T1 &x, T2 b) {
  x += b;
  if (x >= mod) x -= mod;
}
template <typename T1, typename T2>
inline void chmul(T1 &x, T2 b) {
  x = 1LL * x * b % mod;
}
template <typename T1, typename T2>
inline void chmod(T1 &x, T2 b) {
  x %= b, x += b;
  if (x >= b) x -= b;
}
template <typename T>
inline T mabs(T x) {
  return (x < 0 ? -x : x);
}
using namespace std;
using namespace std;
template <typename T>
ostream &operator<<(ostream &cout, vector<T> vec) {
  cout << "{";
  for (int i = 0; i < (int)vec.size(); i++) {
    cout << vec[i];
    if (i != (int)vec.size() - 1) cout << ',';
  }
  cout << "}";
  return cout;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &cout, pair<T1, T2> p) {
  cout << "(" << p.first << ',' << p.second << ")";
  return cout;
}
template <typename T, typename T2>
ostream &operator<<(ostream &cout, set<T, T2> s) {
  vector<T> t;
  for (auto x : s) t.push_back(x);
  cout << t;
  return cout;
}
template <typename T, typename T2>
ostream &operator<<(ostream &cout, multiset<T, T2> s) {
  vector<T> t;
  for (auto x : s) t.push_back(x);
  cout << t;
  return cout;
}
template <typename T>
ostream &operator<<(ostream &cout, queue<T> q) {
  vector<T> t;
  while (q.size()) {
    t.push_back(q.front());
    q.pop();
  }
  cout << t;
  return cout;
}
template <typename T1, typename T2, typename T3>
ostream &operator<<(ostream &cout, map<T1, T2, T3> m) {
  for (auto &x : m) {
    cout << "Key: " << x.first << ' ' << "Value: " << x.second << endl;
  }
  return cout;
}
template <typename T1, typename T2>
void operator+=(pair<T1, T2> &x, const pair<T1, T2> y) {
  x.first += y.first;
  x.second += y.second;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(const pair<T1, T2> &x, const pair<T1, T2> &y) {
  return make_pair(x.first + y.first, x.second + y.second);
}
template <typename T1, typename T2>
pair<T1, T2> operator-(const pair<T1, T2> &x, const pair<T1, T2> &y) {
  return make_pair(x.first - y.first, x.second - y.second);
}
template <typename T1, typename T2>
pair<T1, T2> operator-(pair<T1, T2> x) {
  return make_pair(-x.first, -x.second);
}
template <typename T>
vector<vector<T>> operator~(vector<vector<T>> vec) {
  vector<vector<T>> v;
  int n = vec.size(), m = vec[0].size();
  v.resize(m);
  for (int i = 0; i < m; i++) {
    v[i].resize(n);
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      v[i][j] = vec[j][i];
    }
  }
  return v;
}
void print0x(int x) {
  std::vector<int> vec;
  while (x) {
    vec.push_back(x & 1);
    x >>= 1;
  }
  std::reverse(vec.begin(), vec.end());
  for (int i = 0; i < (int)vec.size(); i++) {
    std::cout << vec[i];
  }
  std::cout << ' ';
}
template <typename T>
void print0x(T x, int len) {
  std::vector<int> vec;
  while (x) {
    vec.push_back(x & 1);
    x >>= 1;
  }
  reverse(vec.begin(), vec.end());
  for (int i = (int)vec.size(); i < len; i++) {
    putchar('0');
  }
  for (size_t i = 0; i < vec.size(); i++) {
    std::cout << vec[i];
  }
  std::cout << ' ';
}
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
namespace combinatorics {
int *fac;
int *ifac;
int __Tmod;
inline int add(int a, int b) { return (a + b) % __Tmod; }
inline int sub(int a, int b) { return (a - b + __Tmod) % __Tmod; }
inline int mult(int a, int b) { return (1LL * a * b) % __Tmod; }
inline int fastpow(int basic, int x) {
  chmod(x, __Tmod - 1);
  if (x == 0) return 1;
  int res = 1;
  while (x) {
    if (x & 1) res = mult(res, basic);
    basic = mult(basic, basic);
    x >>= 1;
  }
  return res;
}
inline int inv(int x) { return fastpow(x, __Tmod - 2); }
inline int div(int a, int b) { return mult(a, inv(b)); }
void init(int n, int tmod) {
  __Tmod = tmod;
  fac = new int[n + 5];
  ifac = new int[n + 5];
  fac[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = mult(fac[i - 1], i);
  }
  ifac[n] = inv(fac[n]);
  for (int i = n - 1; i >= 0; i--) {
    ifac[i] = mult(ifac[i + 1], i + 1);
  }
}
inline int C(int n, int m) {
  if (n < m || n < 0 || m < 0) return 0;
  return mult(mult(fac[n], ifac[m]), ifac[n - m]);
}
inline int Cat(int x) { return mult(C(x * 2, x), inv(x + 1)); }
}  // namespace combinatorics
using namespace std;
using combinatorics::C;
using combinatorics::fac;
int dp[5005][5005];
int n;
int main() {
  combinatorics::init(5000, mod);
  cin >> n;
  dp[0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      chadd(dp[i + 1][j], 1LL * dp[i][j] * j % mod);
      chadd(dp[i + 1][j + 1], 1LL * dp[i][j] * (i + 1 - j) % mod);
    }
  }
  for (int i = 1; i <= n; i++) {
    int res = 0;
    for (int pos = 1; pos <= n; pos++) {
      chadd(res, 1LL * fac[n - pos] * C(n, pos) % mod * dp[pos][i] % mod);
    }
    cout << res << ' ';
  }
  cout << endl;
  return 0;
}