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 maxn = 5e3 + 20;
const int mod = 998244353;
int dp[maxn][maxn], fac[maxn], caf[maxn], ans[maxn];
inline void mkay(int &a) {
  if (a >= mod)
    a -= mod;
  else if (a < 0)
    a += mod;
}
int bpw(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = 1LL * res * a % mod;
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return res;
}
int c(int s, int r) {
  if (r < 0 || r > s) return 0;
  return 1LL * fac[s] * caf[r] % mod * caf[s - r] % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fac[0] = 1;
  for (int i = 1; i < maxn; i++) fac[i] = 1LL * fac[i - 1] * i % mod;
  caf[maxn - 1] = bpw(fac[maxn - 1], mod - 2);
  for (int i = maxn - 1; i > 0; i--) caf[i - 1] = 1LL * caf[i] * i % mod;
  int n;
  cin >> n;
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < i; j++) {
      dp[i][j] = 1LL * (i - j) * dp[i - 1][j] % mod;
      if (j) mkay(dp[i][j] += 1LL * (i - j) * dp[i - 1][j - 1] % mod);
    }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < i; j++)
      dp[i][j] = 1LL * dp[i][j] * c(n, i) % mod * fac[n - i] % mod;
  for (int i = n - 1; i >= 0; i--)
    for (int j = 0; j < i; j++) mkay(dp[i][j] += dp[i + 1][j]);
  for (int i = 0; i < n; i++) {
    int res = 0;
    for (int j = i; j < n; j++) {
      int tmp = 1LL * c(j, i) % mod * dp[j + 1][j] % mod;
      if ((j - i) % 2)
        mkay(res -= tmp);
      else
        mkay(res += tmp);
    }
    cout << res << " ";
  }
  cout << endl;
}