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;
int read() {
  int x = 0;
  bool flg = false;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') flg = true;
  for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
  return flg ? -x : x;
}
const int Mod = 998244353;
inline int qpow(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = 1LL * x * x % Mod)
    if (y & 1) res = 1LL * res * x % Mod;
  return res;
}
inline int upd(int x) { return x + (x >> 31 & Mod); }
int fac[5010], ifac[5010];
int S[5010][5010];
int f[5010];
int n;
int work(int k) {
  int res = 0;
  for (int j = k - 1; j < n; j++) {
    res = upd((res + (j - k + 1 & 1 ? -1LL : 1LL) * fac[j] * ifac[j - k + 1] %
                         Mod * f[j]) %
              Mod);
  }
  res = 1LL * res * fac[n] % Mod * ifac[k - 1] % Mod;
  return res;
}
int main() {
  n = read();
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1LL * fac[i - 1] * i % Mod;
  ifac[n] = qpow(fac[n], Mod - 2);
  for (int i = n; i; i--) ifac[i - 1] = 1LL * ifac[i] * i % Mod;
  S[0][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++)
      S[i][j] = (S[i - 1][j - 1] + 1LL * j * S[i - 1][j]) % Mod;
  for (int j = 0; j < n; j++)
    for (int i = j + 1; i <= n; i++)
      f[j] = (f[j] + 1LL * ifac[i] * S[i][i - j] % Mod * fac[i - j]) % Mod;
  for (int i = 1; i <= n; i++) printf("%d%c", work(i), " \n"[i == n]);
  return 0;
}