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 = 5005, mod = 998244353;
int fac[maxn], inv[maxn];
int o[maxn][maxn];
int read() {
  int x = 0, y = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') y = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * y;
}
int power(int x, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * x % mod;
    x = 1ll * x * x % mod;
    b >>= 1;
  }
  return ans;
}
int C(int n, int m) { return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod; }
int main() {
  int n;
  n = read();
  for (int i = 1; i <= n; i++) {
    o[i][0] = 1;
    o[i][i - 1] = 1;
    for (int k = 1; k < i - 1; k++)
      o[i][k] = (1ll * (k + 1) * o[i - 1][k] % mod +
                 1ll * (i - k) * o[i - 1][k - 1] % mod) %
                mod;
  }
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  inv[n] = power(fac[n], mod - 2);
  for (int i = n - 1; i >= 0; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
  for (int k = 1; k <= n; k++) {
    int ans = 0;
    for (int i = k; i <= n; i++)
      ans = (ans + 1ll * C(n, i) * o[i][k - 1] % mod * fac[n - i] % mod) % mod;
    printf("%d ", ans);
  }
  return 0;
}