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 std::max;
using std::min;
const int inf = 0x3f3f3f3f, Inf = 0x7fffffff;
const long long INF = 0x3f3f3f3f3f3f3f3f;
__inline__ __attribute__((always_inline)) unsigned int rnd() {
  static unsigned int seed = 416;
  return seed ^= seed >> 5, seed ^= seed << 17, seed ^= seed >> 13;
}
template <typename _Tp>
_Tp gcd(const _Tp &a, const _Tp &b) {
  return (!b) ? a : gcd(b, a % b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp abs(const _Tp &a) {
  return a >= 0 ? a : -a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmax(_Tp &a, const _Tp &b) {
  (a < b) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmin(_Tp &a, const _Tp &b) {
  (b < a) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void read(_Tp &x) {
  char ch(getchar());
  bool f(false);
  while (!isdigit(ch)) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (isdigit(ch)) x = (((x << 2) + x) << 1) + (ch & 15), ch = getchar();
  f && (x = -x);
}
template <typename _Tp, typename... Args>
__inline__ __attribute__((always_inline)) void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
template <typename _Tp, typename... Args>
__inline__ __attribute__((always_inline)) _Tp min(const _Tp &a, const _Tp &b,
                                                  const Args &...args) {
  return a < b ? min(a, args...) : min(b, args...);
}
template <typename _Tp, typename... Args>
__inline__ __attribute__((always_inline)) _Tp max(const _Tp &a, const _Tp &b,
                                                  const Args &...args) {
  return a < b ? max(b, args...) : max(a, args...);
}
__inline__ __attribute__((always_inline)) int read_str(char *s) {
  char ch(getchar());
  while (ch == ' ' || ch == '\r' || ch == '\n') ch = getchar();
  char *tar = s;
  *tar = ch, ch = getchar();
  while (ch != ' ' && ch != '\r' && ch != '\n' && ch != EOF)
    *(++tar) = ch, ch = getchar();
  return tar - s + 1;
}
const int N = 5005;
const int mod = 998244353;
template <typename _Tp1, typename _Tp2>
__inline__ __attribute__((always_inline)) void add(_Tp1 &a, _Tp2 b) {
  (a += b) >= mod && (a -= mod);
}
template <typename _Tp1, typename _Tp2>
__inline__ __attribute__((always_inline)) void sub(_Tp1 &a, _Tp2 b) {
  (a -= b) < 0 && (a += mod);
}
int dp[N][N];
long long fac[N], inv[N], ifac[N];
void init() {
  fac[0] = fac[1] = inv[0] = inv[1] = ifac[0] = ifac[1] = 1;
  for (int i = 2; i < N; ++i) {
    fac[i] = fac[i - 1] * i % mod;
    inv[i] = inv[mod % i] * (mod - mod / i) % mod;
    ifac[i] = ifac[i - 1] * inv[i] % mod;
  }
}
__inline__ __attribute__((always_inline)) long long C(int n, int m) {
  if (n < m) return 0;
  return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
int main() {
  init();
  int n;
  read(n);
  dp[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < i; ++j) {
      dp[i][j] = 1ll * dp[i - 1][j] * (j + 1) % mod;
      if (j) add(dp[i][j], 1ll * dp[i - 1][j - 1] * (i - j) % mod);
    }
  }
  for (int k = 0; k < n; ++k) {
    int ans = 0;
    for (int i = 1; i <= n; ++i)
      add(ans, C(n, i) * fac[n - i] % mod * dp[i][k] % mod);
    printf("%d ", ans);
  }
  printf("\n");
  return 0;
}