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;
template <typename T>
inline T read(register T& t) {
  register T f = 1;
  register char ch = getchar();
  t = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') t = t * 10 + ch - '0', ch = getchar();
  t *= f;
  return t;
}
template <typename T, typename... Args>
inline void read(T& t, Args&... args) {
  read(t);
  read(args...);
}
const long long p = 998244353;
inline long long power(register long long x, register long long k = p - 2) {
  register long long re = 1;
  for (; k; k >>= 1, x = x * x % p)
    if (k & 1) re = re * x % p;
  return re;
}
int n;
long long EU[5005][5005];
long long fac[5005], ifac[5005];
int main() {
  read(n);
  fac[0] = 1;
  for (int i = 1; i <= 5000; i++) fac[i] = fac[i - 1] * i % p;
  ifac[5000] = power(fac[5000]);
  for (int i = 5000 - 1; i >= 0; i--) ifac[i] = ifac[i + 1] * (i + 1) % p;
  EU[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    EU[i][0] = 1;
    for (int j = 1; j <= n; j++)
      EU[i][j] = (EU[i - 1][j] * (j + 1) + EU[i - 1][j - 1] * (i - j)) % p;
  }
  for (int i = 1; i <= n; i++) {
    long long ans = 0;
    for (int j = 1; j <= n; j++) ans = (ans + EU[j][i - 1] * ifac[j] % p) % p;
    printf("%lld ", ans * fac[n] % p);
  }
  return 0;
}