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;
int read() {
  int a = 0;
  bool opt = 0;
  char c = getchar();
  while (c < '0' || c > '9') opt |= c == '-', c = getchar();
  while (c >= '0' && c <= '9')
    a = (a << 3) + (a << 1) + (c ^ 48), c = getchar();
  return opt ? -a : a;
}
int n, E[5005][5005], fac[5005], ifac[5005];
int ksm(int a, int n) {
  int ans = 1;
  while (n) {
    if (n & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    n >>= 1;
  }
  return ans;
}
int Mod(int x) { return x < mod ? x : x - mod; }
void Add(int& x, int y) { x = Mod(x + y); }
int main() {
  n = read();
  E[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    E[i][0] = 1;
    for (int j = 1; j < i; ++j) {
      E[i][j] = Mod(1ll * E[i - 1][j] * (j + 1) % mod +
                    1ll * E[i - 1][j - 1] * (i - j) % mod);
    }
  }
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[n] = ksm(fac[n], mod - 2);
  for (int i = n; i; --i) ifac[i - 1] = 1ll * ifac[i] * i % mod;
  for (int i = 1; i <= n; ++i) {
    int ans = 0;
    for (int j = 1; j <= n; ++j) {
      Add(ans, 1ll * E[j][i - 1] * fac[n] % mod * ifac[j] % mod);
    }
    cout << ans << " ";
  }
  return 0;
}