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;
int p1 = 1000000, p2 = 0;
char buf[1000005], wb[1000005];
int gc() {
  if (p1 >= 1000000) fread(buf, 1, 1000000, stdin), p1 = 0;
  return buf[p1++];
}
long long getint() {
  long long ret = 0, flag = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') flag = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    ret = (ret << 3) + (ret << 1) + c - '0';
    c = getchar();
  }
  return ret * flag;
}
void pc(char x) {
  if (p2 >= 1000000) fwrite(wb, 1, 1000000, stdout), p2 = 0;
  wb[p2++] = x;
}
void wrt(long long x) {
  if (x < 0) pc('-'), x = -x;
  int c[24] = {0};
  if (!x) return pc('0'), void();
  while (x) c[++c[0]] = x % 10, x /= 10;
  while (c[0]) pc(c[c[0]--] + '0');
}
int n, m;
long long jac[5005], inv[5005], ans[5005], eu[5005][5005];
long long ksm(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % 998244353;
    a = a * a % 998244353, b >>= 1;
  }
  return ret;
}
int main() {
  n = getint();
  jac[0] = 1, eu[0][0] = 1;
  for (int i = 1; i <= 5000; i++) jac[i] = jac[i - 1] * i % 998244353;
  inv[5000] = ksm(jac[5000], 998244353 - 2);
  for (int i = 5000; i >= 1; i--) inv[i - 1] = inv[i] * i % 998244353;
  for (int i = 1; i <= n; i++) {
    eu[i][0] = 1;
    for (int j = 1; j <= i; j++)
      eu[i][j] = (eu[i - 1][j - 1] * (i - j) % 998244353 +
                  eu[i - 1][j] * (j + 1) % 998244353) %
                 998244353;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++)
      ans[i] =
          (ans[i] + eu[j][i - 1] * jac[n] % 998244353 * inv[j] % 998244353) %
          998244353;
  }
  for (int i = 1; i <= n; i++) wrt(ans[i]), pc(' ');
  fwrite(wb, 1, p2, stdout);
  return 0;
}