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>
namespace IO {
char gc() { return getchar(); }
template <typename Tp>
bool get1(Tp &x) {
  bool neg = 0;
  char c = gc();
  while (c != EOF && (c < '0' || c > '9') && c != '-') c = gc();
  if (c == '-') c = gc(), neg = 1;
  if (c == EOF) return false;
  x = 0;
  for (; c >= '0' && c <= '9'; c = gc()) x = x * 10 + c - '0';
  if (neg) x = -x;
  return true;
}
template <typename Tp>
void printendl(Tp x) {
  if (x < 0) putchar('-'), x = -x;
  static short a[40], sz;
  sz = 0;
  while (x > 0) a[sz++] = x % 10, x /= 10;
  if (sz == 0) putchar('0');
  for (int i = sz - 1; i >= 0; i--) putchar('0' + a[i]);
  puts("");
}
}  // namespace IO
using IO::get1;
using IO::printendl;
const int inf = 0x3f3f3f3f;
const long long Linf = 1ll << 61;
const int maxn = 5111;
const int mod = 998244353;
int n, E[maxn][maxn], C[maxn][maxn], fac[maxn];
int main() {
  get1(n);
  E[1][0] = 1;
  for (int i = 2; i <= n; i++)
    for (int j = 0; j < i; j++) {
      E[i][j] = 1ll * E[i - 1][j] * (j + 1) % mod;
      if (j) E[i][j] = (E[i][j] + 1ll * E[i - 1][j - 1] * (i - j)) % mod;
    }
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * i * fac[i - 1] % mod;
  for (int i = 0; i <= n; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
      if (C[i][j] >= mod) C[i][j] -= mod;
    }
  }
  for (int i = 0; i < n; i++) {
    int sum = 0;
    for (int j = 1; j <= n; j++)
      sum = (sum + 1ll * E[j][i] * fac[n - j] % mod * C[n][j]) % mod;
    printf("%d ", sum);
  }
  return 0;
}