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>
const int N = 5010, mod = 998244353;
int n;
struct z {
  int x;
  z(int x = 0) : x(x) {}
  friend inline z operator*(z a, z b) { return (long long)a.x * b.x % mod; }
  friend inline z operator-(z a, z b) {
    return (a.x -= b.x) < 0 ? a.x + mod : a.x;
  }
  friend inline z operator+(z a, z b) {
    return (a.x += b.x) >= mod ? a.x - mod : a.x;
  }
} ans[N], fac[N], inv[N], ifac[N], g[N][N];
inline z fpow(z a, int b) {
  z s = 1;
  for (; b; b >>= 1, a = a * a)
    if (b & 1) s = s * a;
  return s;
}
void init(int n) {
  fac[0] = ifac[0] = inv[0] = inv[1] = 1;
  for (int i = 2; i <= n; i++) inv[i] = (mod - mod / i) * inv[mod % i];
  for (int i = 1; i <= n; i++)
    fac[i] = fac[i - 1] * i, ifac[i] = ifac[i - 1] * inv[i];
}
int main() {
  std::cin >> n;
  init(n + 5);
  g[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      g[i][j] = g[i - 1][j] * j + g[i - 1][j - 1];
      ans[i - j] = ans[i - j] + g[i][j] * fac[j] * fac[n] * ifac[i];
    }
  }
  for (int i = n - 1; ~i; i--) {
    for (int j = i + 1; j < n; j++)
      ans[i] = ans[i] - fac[j] * ifac[i] * ifac[j - i] * ans[j];
  }
  for (int i = 0; i < n; i++) std::cout << ans[i].x << " ";
  puts("");
}