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 N = 5005;
template <typename T>
inline void read(T &AKNOI) {
  T x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  AKNOI = flag * x;
}
namespace ModCalculator {
const int MOD = 998244353;
inline void Inc(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
inline void Dec(int &x, int y) {
  x -= y;
  if (x < 0) x += MOD;
}
inline int Add(int x, int y) {
  Inc(x, y);
  return x;
}
inline int Sub(int x, int y) {
  Dec(x, y);
  return x;
}
inline int Mul(int x, int y) { return 1LL * x * y % MOD; }
inline int Ksm(int x, int k) {
  int ret = 1;
  for (; k; k >>= 1) {
    if (k & 1) ret = Mul(ret, x);
    x = Mul(x, x);
  }
  return ret;
}
inline int Inv(int x) { return Ksm(x, MOD - 2); }
}  // namespace ModCalculator
using namespace ModCalculator;
int n, fac[N], finv[N], E[N][N];
void init() {
  read(n);
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) {
    fac[i] = Mul(fac[i - 1], i);
  }
  finv[n] = Inv(fac[n]);
  for (int i = n; i >= 1; --i) {
    finv[i - 1] = Mul(finv[i], i);
  }
  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(Mul(E[i - 1][j], j + 1), Mul(E[i - 1][j - 1], i - j));
    }
  }
}
void solve() {
  for (int i = 1; i <= n; ++i) {
    int ans = 0;
    for (int j = i; j <= n; ++j) {
      Inc(ans, Mul(E[j][i - 1], Mul(fac[n], finv[j])));
    }
    printf("%d ", ans);
  }
}
int main() {
  init();
  solve();
  return 0;
}