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;
long long n, fac[5005], ifac[5005], inv[5005], C[5005][5005], ans[5005];
int main() {
  scanf("%lld", &n);
  fac[0] = fac[1] = ifac[0] = ifac[1] = inv[0] = inv[1] = 1;
  for (long long i = 2; i <= n; i++) {
    fac[i] = fac[i - 1] * i % 998244353;
    inv[i] = (998244353 - 998244353 / i) * inv[998244353 % i] % 998244353;
    ifac[i] = ifac[i - 1] * inv[i] % 998244353;
  }
  for (long long i = 1; i <= n; i++) {
    C[i][0] = 1;
    for (long long t = 1; t <= i; t++) {
      C[i][t] = ((i - t) * C[i - 1][t - 1] + (t + 1) * C[i - 1][t]) % 998244353;
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long t = 1; t <= n; t++) {
      ans[i] =
          (ans[i] + C[t][i - 1] * fac[n] % 998244353 * ifac[t]) % 998244353;
    }
  }
  for (long long i = 1; i <= n; i++) printf("%lld ", ans[i]);
}