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 int mod = 998244353;
long long S[5010][5010], tmp[5010];
long long sum[5010];
long long fac[5010], inv[5010];
long long power(int a, int b) {
  if (!b) return 1;
  long long tmp = power(a, b >> 1);
  return b & 1 ? tmp * tmp % mod * a % mod : tmp * tmp % mod;
}
long long C(int n, int k) { return fac[n] * inv[k] % mod * inv[n - k] % mod; }
int main() {
  fac[0] = 1, inv[0] = 1;
  for (int i = 1; i <= 5005; i++) {
    fac[i] = fac[i - 1] * i % mod;
    inv[i] = power(fac[i], mod - 2);
  }
  int n;
  scanf("%d", &n);
  S[1][1] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = i; j >= 1; j--) {
      S[i][j] = (S[i - 1][j - 1] + S[i - 1][j] * j) % mod;
    }
  }
  for (int i = 1; i <= n; i++) {
    reverse(S[i] + 1, S[i] + 1 + i);
    for (int j = 0; j < i; j++) {
      S[i][j] = S[i][j + 1] * fac[i - j] % mod;
    }
  }
  for (int j = 0; j < n; j++) {
    for (int i = n; i >= 0; i--) {
      S[i][j] = S[i][j] * fac[n] % mod * inv[i] % mod;
      S[i][j] = (S[i][j] + S[i + 1][j]) % mod;
    }
  }
  long long mul = 1;
  for (int i = n; i >= 1; i--) {
    sum[i] = (sum[i + 1] + mul) % mod;
    mul = mul * i % mod;
  }
  for (int i = 0; i < n; i++) {
    long long ans = 0;
    for (int j = i; j < n; j++) {
      ans = (ans + C(j, i) * S[j + 1][j] % mod * (j - i & 1 ? -1 : 1) + mod) %
            mod;
    }
    printf("%I64d ", ans);
  }
  printf("\n");
  return 0;
}