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 = 5e3 + 51, MOD = 998244353;
int n, res;
int p[MAXN][MAXN], fact[MAXN], finv[MAXN];
inline int read() {
  register int num = 0, neg = 1;
  register char ch = getchar();
  while (!isdigit(ch) && ch != '-') {
    ch = getchar();
  }
  if (ch == '-') {
    neg = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    num = (num << 3) + (num << 1) + (ch - '0');
    ch = getchar();
  }
  return num * neg;
}
inline int qpow(int base, int exponent) {
  int res = 1;
  while (exponent) {
    if (exponent & 1) {
      res = (long long int)res * base % MOD;
    }
    base = (long long int)base * base % MOD, exponent >>= 1;
  }
  return res;
}
inline void setup(int cnt) {
  fact[0] = fact[1] = finv[0] = 1;
  for (register int i = 0; i <= cnt; i++) {
    p[i][0] = 1;
  }
  for (register int i = 1; i <= cnt; i++) {
    for (register int j = 1; j <= i; j++) {
      p[i][j] = ((long long int)(i - j) * p[i - 1][j - 1] % MOD +
                 (long long int)(j + 1) * p[i - 1][j] % MOD) %
                MOD;
    }
  }
  for (register int i = 2; i <= cnt; i++) {
    fact[i] = (long long int)fact[i - 1] * i % MOD;
  }
  finv[cnt] = qpow(fact[cnt], MOD - 2);
  for (register int i = cnt - 1; i; i--) {
    finv[i] = (long long int)finv[i + 1] * (i + 1) % MOD;
  }
}
inline int binom(int m, int n) {
  return (long long int)fact[m] * finv[n] % MOD * finv[m - n] % MOD;
}
int main() {
  setup(n = read());
  for (register int k = 0; k < n; k++) {
    res = 0;
    for (register int i = max(k, 1); i <= n; i++) {
      res = (res +
             (long long int)binom(n, i) * p[i][k] % MOD * fact[n - i] % MOD) %
            MOD;
    }
    printf("%d ", res);
  }
  puts("");
  return 0;
}