PolandBall is standing in a row with Many Other Balls. More precisely, there are exactly n Balls. Balls are proud of their home land — and they want to prove that it's strong.

The Balls decided to start with selecting exactly m groups of Balls, each consisting either of single Ball or two neighboring Balls. Each Ball can join no more than one group.

The Balls really want to impress their Enemies. They kindly asked you to calculate number of such divisions for all m where 1 ≤ m ≤ k. Output all these values modulo 998244353, the Enemies will be impressed anyway.

Input

There are exactly two numbers n and k (1 ≤ n ≤ 109, 1 ≤ k < 215), denoting the number of Balls and the maximim number of groups, respectively.

Output

You should output a sequence of k values. The i-th of them should represent the sought number of divisions into exactly i groups, according to PolandBall's rules.

Examples

Input

3 3


Output

5 5 1 

Input

1 1


Output

1 

Input

5 10


Output

9 25 25 9 1 0 0 0 0 0 

Note

In the first sample case we can divide Balls into groups as follows: 

{1}, {2}, {3}, {12}, {23}.

{12}{3}, {1}{23}, {1}{2}, {1}{3}, {2}{3}.

{1}{2}{3}.

Therefore, output is: 5 5 1.
#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int N = 1 << 15;
int Jie[N], inv[N], jie[N];
inline int mul(const int &a, const int &b) { return 1ll * a * b % mod; }
inline void add(int &a, const int &b) {
  a += b;
  if (a >= mod) a -= mod;
  return;
}
inline int C(int m, int n) { return mul(jie[m], mul(inv[n], inv[m - n])); }
inline int CC(int m, int n) { return mul(Jie[m], inv[n]); }
int main() {
  int n = 1000000000, k = 32767;
  scanf("%d%d", &n, &k);
  Jie[0] = n;
  jie[0] = 1, inv[0] = inv[1] = 1;
  for (int i = (2); i < (k + 1); ++i)
    inv[i] = -1ll * mod / i * inv[mod % i] % mod + mod;
  for (int i = (1); i < (k + 1); ++i) {
    Jie[i] = n - i;
    jie[i] = mul(jie[i - 1], i);
    inv[i] = mul(inv[i], inv[i - 1]);
  }
  for (int i = (1); i < (min(k, n) + 1); ++i) {
    int ans = 0;
    for (int j = (0); j < (i + 1); ++j) {
      if (n - j < i) break;
      ans = (ans + 1ll * C(i, j) * CC(j, i)) % mod;
    }
    for (int j = (0); j < (k + 1); ++j)
      Jie[j] = 1ll * Jie[j] * (n - i - j) % mod;
    printf("%d ", ans);
  }
  for (int i = (min(k, n) + 1); i < (k + 1); ++i) {
    printf("0 ");
  }
}
