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 double pi = acos(-1);
int mod = 998244353;
long long Mod(long long x) { return x >= mod ? x - mod : x; }
long long read() {
  long long x = 0;
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + (ch ^ 48);
  return x * f;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
long long qpow(long long a, int b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return ans;
}
const int N = 5010;
long long fac[N], ifac[N];
long long C(int n, int m) { return fac[n] * ifac[m] % mod * ifac[n - m] % mod; }
long long p[N][N];
void init(int n) {
  p[0][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++)
      p[i][j] = (j * p[i - 1][j] + (i - j + 1) * p[i - 1][j - 1]) % mod;
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % mod;
  ifac[n] = qpow(fac[n], mod - 2);
  for (int i = n; i; i--) ifac[i - 1] = ifac[i] * i % mod;
}
int n;
long long ans;
int main() {
  init(n = read());
  for (int k = 1; k <= n; k++) {
    ans = 0;
    for (int i = k; i <= n; i++)
      ans = (ans + C(n, i) * p[i][k] % mod * fac[n - i]) % mod;
    print(ans), putchar(' ');
  }
}