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>
#pragma comment(linker, "/STACK:1000000000")
#pragma GCC optimize("Ofast,no-stack-protector")
using namespace std;
const int mod = 998244353;
int add(int x, int y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
  return x;
}
int sub(int x, int y) {
  x -= y;
  if (x < 0) {
    x += mod;
  }
  return x;
}
int mul(int x, int y) { return (long long)x * y % mod; }
int my_pow(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = mul(x, x)) {
    if (y & 1) {
      ans = mul(ans, x);
    }
  }
  return ans;
}
const int maxn = 5010;
int dp[maxn][maxn];
int f[maxn], rf[maxn];
int main() {
  f[0] = 1;
  for (int i = 1; i < maxn; i++) {
    f[i] = mul(i, f[i - 1]);
    rf[i] = my_pow(f[i], mod - 2);
  }
  int n;
  cin >> n;
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    int s = 0;
    for (int j = 1; j <= i; j++) {
      dp[i][j] = add(mul(dp[i - 1][j], j), mul(dp[i - 1][j - 1], i - j + 1));
      s = add(s, dp[i][j]);
    }
  }
  for (int x = 0; x < n; x++) {
    int ans = 0;
    for (int i = x + 1; i <= n; i++) {
      ans = add(ans, mul(dp[i][x + 1], rf[i]));
    }
    ans = mul(ans, f[n]);
    printf("%d ", ans);
  }
}