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 N = 5010;
const int mod = 998244353;
int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
int qpow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return ret;
}
int n, d[N][N], S[N][N], fac[N], f[N], ifac[N];
int main() {
  fac[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[N - 1] = qpow(fac[N - 1], mod - 2);
  for (int i = N - 1; 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 * S[i - 1][j] * j) % mod;
  for (int i = 0; i < N; i++)
    for (int j = 0; j <= i; j++)
      d[i][j] = 1ll * S[i + 1][i + 1 - j] * fac[i + 1 - j] % mod;
  cin >> n;
  for (int y = 0; y < n; y++)
    for (int x = y; x < n; x++)
      f[y] = (f[y] + 1ll * d[x][y] * ifac[x + 1]) % mod;
  for (int i = 0; i < n; i++) {
    int ans = 0;
    for (int y = i; y < n; y++)
      ans = (ans + 1ll * fac[y] * ((y - i) & 1 ? mod - 1 : 1) % mod *
                       ifac[y - i] % mod * f[y]) %
            mod;
    ans = 1ll * ans * fac[n] % mod * ifac[i] % mod;
    cout << ans << ' ';
  }
  return 0;
}