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;
inline void read(int &x) {
  int v = 0, f = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = (c & 15);
  while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
  x = v * f;
}
inline void read(long long &x) {
  long long v = 0ll, f = 1ll;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = (c & 15);
  while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
  x = v * f;
}
inline void readc(char &x) {
  char c;
  while (((c = getchar()) == ' ') || c == '\n')
    ;
  x = c;
}
const int mod = 998244353;
int n, m, i, j, f[5005][5005], c[5005][5005], fac[5005], ans[5005];
int main() {
  read(n);
  f[0][0] = c[0][0] = fac[0] = 1;
  for (((i)) = (1); ((i)) <= ((n)); ((i))++) {
    fac[i] = 1ll * fac[i - 1] * i % mod;
    c[i][0] = 1;
    for (((j)) = (1); ((j)) <= ((i)); ((j))++) {
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
    }
    for (((j)) = (0); ((j)) <= (((int)(i)) - 1); ((j))++) {
      f[i][j] =
          (1ll * (j + 1) * f[i - 1][j] + 1ll * (i - j) * f[i - 1][j - 1]) % mod;
    }
  }
  for (((i)) = (0); ((i)) <= (((int)(n)) - 1); ((i))++) {
    for ((j) = (i + 1); (j) <= (n); (j)++) {
      ans[i] = (ans[i] + 1ll * f[j][i] * c[n][j] % mod * fac[n - j]) % mod;
    }
  }
  for (((i)) = (0); ((i)) <= (((int)(n)) - 1); ((i))++) printf("%d ", ans[i]);
  return 0;
}