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>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
const long long MOD2 = (long long)998244353 * (long long)998244353;
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 998244353) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 998244353) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
int dp[5035][5035], fac[5035], C[5035][5035];
void init(int mod = 998244353) {
  C[0][0] = 1;
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      C[i][j] = C[i - 1][j - 1];
      addmod(C[i][j], C[i - 1][j]);
    }
  }
  fac[0] = 1;
  for (int(i) = 1; (i) <= (int)(n); (i)++)
    fac[i] = (long long)fac[i - 1] * i % 998244353;
  dp[1][0] = 1;
  for (int(i) = 1; (i) <= (int)(n - 1); (i)++) {
    for (int j = 0; j < i; j++) {
      addmod(dp[i + 1][j], dp[i][j]);
      addmod(dp[i + 1][j + 1], dp[i][j]);
      addmod(dp[i + 1][j], (long long)dp[i][j] * j % 998244353);
      addmod(dp[i + 1][j + 1], (long long)dp[i][j] * (i - 1 - j) % 998244353);
    }
  }
}
void fmain(int tid) {
  scanf("%d", &n);
  init();
  for (int(k) = 1; (k) <= (int)(n); (k)++) {
    int ans = 0;
    for (int(i) = 1; (i) <= (int)(n); (i)++) {
      addmod(ans, (long long)dp[i][k - 1] * C[n][i] % 998244353 * fac[n - i] %
                      998244353);
    }
    printf("%d ", ans);
  }
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}