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]\}.
#include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;
const int N = 5e3 + 10;
const long long mod2 = 1e9 + 7;
const long long mod = 998244353;
const long long inf = 8e18;
const int LOG = 22;
long long pw(long long a, long long b, long long M) {
  return (!b ? 1
             : (b & 1 ? (a * pw(a * a % M, b / 2, M)) % M
                      : pw(a * a % M, b / 2, M)));
}
long long dp[N][N], Ans[N], first[N], I[N];
long long E(long long r, long long n) {
  if (r > n || r < 0) return 0ll;
  return (first[n] * I[r] % mod) * I[n - r] % mod;
}
int main() {
  first[0] = 1;
  for (int i = 1; i < N; i++) first[i] = (i * first[i - 1]) % mod;
  I[N - 1] = pw(first[N - 1], mod - 2, mod);
  for (int i = N - 2; ~i; i--) I[i] = I[i + 1] * (i + 1) % mod;
  int n;
  scanf("%d", &n);
  dp[1][1] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      dp[i][j] =
          ((dp[i - 1][j] * j % mod) + (dp[i - 1][j - 1] * (i - j + 1) % mod)) %
          mod;
    }
  }
  for (int i = 1; i <= n; i++) {
    Ans[i] = 0;
    for (int j = 1; j <= n; j++) {
      Ans[i] = (Ans[i] + (first[n] * I[j] % mod) * dp[j][i]) % mod;
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%lld ", Ans[i]);
  }
  return 0;
}
