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 maxn = 5000;
const int mod = 998244353;
int n;
int E[maxn + 5][maxn + 5];
int ans;
namespace usual {
int fac[maxn + 5], ifac[maxn + 5];
int add(int ta, int tb) { return ta + tb >= mod ? ta + tb - mod : ta + tb; }
int sub(int ta, int tb) { return ta < tb ? ta - tb + mod : ta - tb; }
int ksm(long long ta, int tp) {
  int s = 1;
  for (; tp; ta = ta * ta % mod, tp >>= 1)
    if (tp & 1) s = ta * s % mod;
  return s;
}
void make_fac() {
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * i * fac[i - 1] % mod;
  ifac[n] = ksm(fac[n], mod - 2);
  for (int i = n; i; i--) ifac[i - 1] = 1ll * ifac[i] * i % mod;
}
}  // namespace usual
using namespace usual;
void make_E() {
  E[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    E[i][0] = 1;
    for (int j = 1; j < i; j++)
      E[i][j] = add(1ll * (j + 1) * E[i - 1][j] % mod,
                    1ll * (i - j) * E[i - 1][j - 1] % mod);
  }
}
int main() {
  scanf("%d", &n);
  make_fac();
  make_E();
  for (int i = 1; i <= n; i++) {
    ans = 0;
    for (int j = 1; j <= n; j++)
      ans = add(ans, 1ll * E[j][i - 1] * fac[n] % mod * ifac[j] % mod);
    printf("%d ", ans);
  }
  printf("\n");
  return 0;
}