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 JAVA):

Solution:

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

public final class F {

    private static final int MOD = 998244353;
    private static final int N = (int) (5e3 + 5);

    public static void main(String[] args) {
        final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        final int n = in.nextInt();
        final long[][] dp = new long[N][N];
        dp[0][0] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j <= i; ++j) {
                if (j > 0) {
                    dp[i][j] = (dp[i][j] + (i - j) * dp[i - 1][j - 1] % MOD) % MOD;
                }
                dp[i][j] = (dp[i][j] + (j + 1) * dp[i - 1][j] % MOD) % MOD;
            }
        }
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j <= i; ++j) {
                if (j > 0) {
                    dp[i][j] = (dp[i][j] + i * dp[i - 1][j - 1] % MOD) % MOD;
                }
            }
        }
        for (int i = n - 1; i >= 0; --i) {
            System.out.print(dp[n][i] + " ");
        }
        System.out.println();
    }
}