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 N = 5010;
const int mod = 998244353;
inline int add(int a, int b) {
  a += b;
  return a >= mod ? a - mod : a;
}
inline int sub(int a, int b) {
  a -= b;
  return a < 0 ? a + mod : a;
}
inline int mul(int a, int b) { return 1ll * a * b % mod; }
inline int qpow(int a, int b) {
  int ret = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) ret = mul(ret, a);
  return ret;
}
int g[N][N];
int fac[N], ifac[N];
void init(int n = 5000) {
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = mul(fac[i - 1], i);
  ifac[n] = qpow(fac[n], mod - 2);
  for (int i = n - 1; i; i--) ifac[i] = mul(ifac[i + 1], i + 1);
}
int ans[N], n;
int main() {
  init();
  cin >> n;
  g[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      g[i][j] = add(mul(g[i - 1][j], j), mul(i - (j - 1), g[i - 1][j - 1]));
      ans[j] = add(ans[j], mul(g[i][j], mul(fac[n], ifac[i])));
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  puts("");
}