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 long long mod = 998244353;
const long long N = 5e3 + 100;
long long fact[N + 1], ifact[N + 1], d[N + 1][N + 1], ans[N + 1];
long long power(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  if (b % 2 == 0) {
    long long t = power(a, b / 2);
    return t * t % mod;
  }
  return a * power(a, b - 1) % mod;
}
signed main() {
  long long n;
  cin >> n;
  d[0][0] = fact[0] = ifact[0] = 1;
  for (long long i = 1; i <= N; i++) {
    fact[i] = fact[i - 1] * i % mod;
  }
  ifact[N] = power(fact[N], mod - 2);
  for (long long i = N - 1; i >= 0; i--) {
    ifact[i] = ifact[i + 1] * (i + 1) % mod;
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= i; j++) {
      d[i][j] = (d[i - 1][j - 1] * (i - j + 1) + j * d[i - 1][j]) % mod;
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      ans[i] = (ans[i] + d[j][i] * ifact[j]) % mod;
    }
  }
  for (long long i = 1; i <= n; i++) {
    cout << ans[i] * fact[n] % mod << " ";
  }
  return 0;
}