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 P = 998244353, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 5e3 + 10;
int n, a[N], fac[N], T[N][N], C[N][N];
int solve(int k) {
  int ans = 0;
  for (int i = k - 1; i <= n; ++i)
    ans = (ans + (long long)T[i][k - 1] * C[n][i] % P * fac[n - i]) % P;
  return ans;
}
int main() {
  scanf("%d", &n);
  C[0][0] = fac[0] = 1;
  for (int i = 1; i <= n; ++i) {
    fac[i] = (long long)fac[i - 1] * i % P;
    C[i][0] = T[i][0] = 1;
    for (int j = 1; j <= i; ++j) {
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % P;
      T[i][j] =
          ((j + 1ll) * T[i - 1][j] + (long long)(i - j) * T[i - 1][j - 1]) % P;
    }
  }
  for (int i = 1; i <= n; ++i) printf("%d ", solve(i));
  putchar(10);
}