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 kMod = 998244353;
const int kN = 5e3 + 10;
int fpow(int a, int n) {
  int res = 1;
  while (n) {
    if (n & 1) res = 1LL * res * a % kMod;
    n >>= 1, a = 1LL * a * a % kMod;
  }
  return res;
}
int Inv(int a) { return fpow(a, kMod - 2); }
int n;
int fac[kN], invf[kN];
void pre() {
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1LL * fac[i - 1] * i % kMod;
  invf[n] = Inv(fac[n]);
  for (int i = n - 1; i >= 0; i--) invf[i] = 1LL * invf[i + 1] * (i + 1) % kMod;
}
int C(int n, int m) {
  if (n < m) return 0;
  return 1LL * fac[n] * invf[m] % kMod * invf[n - m] % kMod;
}
int dp[kN][kN];
int f[kN];
int add(int a, int b) {
  a += b;
  if (a >= kMod) a -= kMod;
  return a;
}
int main() {
  scanf("%d", &n);
  pre();
  dp[0][0] = 1;
  for (int k = 1; k <= n; k++) {
    for (int i = k; i <= n; i++)
      dp[i][k] = add(1LL * k * dp[i - 1][k] % kMod,
                     1LL * (i - k + 1) * dp[i - 1][k - 1] % kMod);
  }
  for (int i = 1; i <= n; i++) {
    int ans = 0;
    for (int j = i; j <= n; j++)
      ans = add(ans, 1LL * fac[n] * invf[j] % kMod * dp[j][i] % kMod);
    printf("%d ", ans);
  }
  return 0;
}