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>
const int CN = 5050, P = 998244353;
int n, fac[CN], ifac[CN], E[CN][CN];
int C(int n, int m) {
  return m < 0 || m > n ? 0 : (long long)fac[n] * ifac[m] % P * ifac[n - m] % P;
}
int main() {
  E[0][0] = 1;
  for (int i = 1; i <= 5000 && (E[i][0] = 1); i++)
    for (int j = 1; j <= i; j++)
      E[i][j] = ((long long)(j + 1) * E[i - 1][j] +
                 (long long)(i - j) * E[i - 1][j - 1]) %
                P;
  fac[0] = ifac[0] = ifac[1] = 1;
  for (int i = 1; i <= 5000; i++) fac[i] = (long long)fac[i - 1] * i % P;
  for (int i = 2; i <= 5000; i++)
    ifac[i] = (long long)ifac[P % i] * (P - P / i) % P;
  for (int i = 2; i <= 5000; i++)
    ifac[i] = (long long)ifac[i - 1] * ifac[i] % P;
  scanf("%d", &n);
  for (int i = 1, res = 0; i <= n; res = 0, i++) {
    for (int j = 1; j <= n; j++)
      res = (res + (long long)E[j][i - 1] * C(n, j) % P * fac[n - j]) % P;
    printf("%d ", res);
  }
  puts("");
  return 0;
}