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 <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const int N = 5e3 + 5, mod = 998244353;
long long dp[N][N], fac[N], inv[N], ans[N];
long long pw(long long a, long long b, long long m = mod) {
  long long ret = 1;
  while (b > 0) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  dp[0][0] = fac[0] = inv[0] = 1;
  for (int i = (1); ((1) > 0 ? i < (n + 1) : i > (n + 1)); i += (1))
    fac[i] = fac[i - 1] * i % mod;
  inv[n] = pw(fac[n], mod - 2);
  for (int i = (n - 1); ((-1) > 0 ? i < (0) : i > (0)); i += (-1))
    inv[i] = inv[i + 1] * (i + 1) % mod;
  for (int i = (1); ((1) > 0 ? i < (n + 1) : i > (n + 1)); i += (1)) {
    for (int j = (1); ((1) > 0 ? j < (i + 1) : j > (i + 1)); j += (1)) {
      dp[i][j] = (dp[i - 1][j - 1] * (i - j + 1) + dp[i - 1][j] * j) % mod;
    }
  }
  for (int i = (1); ((1) > 0 ? i < (n + 1) : i > (n + 1)); i += (1)) {
    for (int j = (1); ((1) > 0 ? j < (n + 1) : j > (n + 1)); j += (1)) {
      ans[i] = (ans[i] + dp[j][i] * inv[j]) % mod;
    }
  }
  for (int i = (1); ((1) > 0 ? i < (n + 1) : i > (n + 1)); i += (1))
    cout << ans[i] * fac[n] % mod << " \n"[i == n];
}