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>
int o[5050][5050], ans[5050], ifac[5050], fac[5050], n;
inline int quickpow(int a, int b) {
  register int res = 1;
  while (b) {
    if (b & 1) res = 1LL * res * a % 998244353;
    a = 1LL * a * a % 998244353;
    b >>= 1;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  o[0][0] = 1;
  for (register int i = 1; i <= n; ++i) {
    o[i][0] = 1;
    for (register int j = 1; j < i; ++j)
      o[i][j] =
          (1LL * (j + 1) * o[i - 1][j] + 1LL * (i - j) * o[i - 1][j - 1]) %
          998244353;
  }
  fac[0] = ifac[0] = 1;
  for (register int i = 1; i <= n; ++i)
    fac[i] = 1LL * fac[i - 1] * i % 998244353;
  ifac[n] = quickpow(fac[n], 998244353 - 2);
  for (register int i = n - 1; i; --i)
    ifac[i] = 1LL * ifac[i + 1] * (i + 1) % 998244353;
  for (register int i = 1; i <= n; ++i)
    for (register int j = 1; j <= n; ++j)
      if (o[j][i - 1])
        ans[i] = (ans[i] + 1LL * o[j][i - 1] * ifac[j]) % 998244353;
  for (register int i = 1; i <= n; ++i) {
    ans[i] = 1LL * ans[i] * fac[n] % 998244353;
    printf("%d ", ans[i]);
  }
}