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 int gi() {
  int f = 1, sum = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    sum = (sum << 3) + (sum << 1) + ch - '0';
    ch = getchar();
  }
  return f * sum;
}
const int N = 5010, Mod = 998244353;
int n, a[N][N], c[N][N];
int main() {
  n = gi();
  c[1][1] = c[2][1] = c[2][2] = 1;
  for (int i = 3; i <= n; i++) {
    int l = i - 1, r = 2;
    c[i][1] = 1;
    for (int j = 2; j <= n; j++) {
      c[i][j] =
          (1ll * c[i - 1][j - 1] * l % Mod + 1ll * c[i - 1][j] * r % Mod) % Mod;
      l--;
      r++;
    }
  }
  a[1][1] = 1;
  for (int i = 2; i <= n; i++) {
    a[i][i] = 1;
    for (int j = 1; j < i; j++) {
      a[i][j] = (1ll * a[i - 1][j] * i % Mod + c[i][j]) % Mod;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d%c", a[n][i], " \n"[i == n]);
  return 0;
}