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 inf = 2e9;
const long long INF = 8e18;
const int maxn = 5e3 + 5;
const int mod = 998244353;
int n;
long long dp[maxn][maxn], fac[maxn], inv[maxn];
inline long long powmod(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
inline void init() {
  dp[0][0] = fac[0] = inv[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = fac[i - 1] * i % mod;
  inv[n] = powmod(fac[n], mod - 2);
  for (int i = n - 1; i >= 1; --i) inv[i] = inv[i + 1] * (i + 1) % mod;
  for (int i = 1; i <= n; ++i) {
    dp[i][0] = 1;
    for (int j = 1; j <= i - 1; ++j) {
      dp[i][j] =
          dp[i - 1][j] * (j + 1) % mod + dp[i - 1][j - 1] * (i - j) % mod;
      dp[i][j] %= mod;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  init();
  for (int i = 1; i <= n; ++i) {
    long long ans = 0;
    for (int j = i; j <= n; ++j) {
      ans += dp[j][i - 1] * inv[j] % mod;
    }
    ans %= mod;
    ans *= fac[n];
    ans %= mod;
    cout << ans << ' ';
  }
  return 0;
}