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 mod = 998244353;
inline int add(int a, int b) {
  if ((a += b) >= mod) a -= mod;
  return a;
}
inline int dec(int a, int b) {
  if ((a -= b) < 0) a += mod;
  return a;
}
inline int mult(int a, int b) {
  long long t = 1ll * a * b;
  if (t >= mod) t %= mod;
  return t;
}
inline int power(int a, int b) {
  int out = 1;
  while (b) {
    if (b & 1) out = mult(out, a);
    a = mult(a, a);
    b >>= 1;
  }
  return out;
}
int n, d[5010][5010], fac[5010], ifac[5010], ans[5010];
int main() {
  scanf("%d", &n);
  fac[0] = 1;
  d[0][0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = mult(fac[i - 1], i);
  ifac[n] = power(fac[n], mod - 2);
  for (int i = n - 1; i >= 0; i--) ifac[i] = mult(ifac[i + 1], i + 1);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      d[i][j] = add(mult(d[i - 1][j - 1], i - j + 1), mult(d[i - 1][j], j));
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      ans[i] = add(ans[i], mult(d[j][i], ifac[j]));
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", mult(ans[i], fac[n]));
  return 0;
}