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;
template <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int cys = 998244353;
int n;
long long fac[5005], inv[5005], d[5005][5005], ans[5005];
long long qpow(long long x, long long p) {
  long long ret = 1;
  for (; p; p >>= 1, x = x * x % cys)
    if (p & 1) ret = ret * x % cys;
  return ret;
}
int main() {
  n = readint();
  d[0][0] = fac[0] = inv[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % cys;
  inv[n] = qpow(fac[n], cys - 2);
  for (int i = n - 1; i >= 1; i--) inv[i] = inv[i + 1] * (i + 1) % cys;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++)
      d[i][j] = (d[i - 1][j - 1] * (i - j + 1) + d[i - 1][j] * j) % cys;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) ans[i] = (ans[i] + d[j][i] * inv[j]) % cys;
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i] * fac[n] % cys);
  printf("\n");
  return 0;
}