Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = (1 << 16) + 5, P = 998244353;
int n, K;
void m1(int &x) {
  if (x >= P) x -= P;
}
int ksm(int a, int b) {
  int x = 1;
  for (; b; b & 1 ? x = 1ll * x * a % P : 0, a = 1ll * a * a % P, b >>= 1)
    ;
  return x;
}
namespace poly {
const int WN = 3;
int len;
int tr[N], w[N], A[N], B[N], C[N], D[N];
void pret(int l) {
  for (len = 1; len < l; len <<= 1)
    ;
  for (int i = 0; i < len; i++)
    tr[i] = (tr[i >> 1] >> 1) | (i & 1 ? len >> 1 : 0);
}
void prew(int l) {
  for (len = 1; len < l; len <<= 1)
    ;
  for (int mid = 1; mid < len; mid <<= 1) {
    int wn = ksm(WN, (P - 1) / (2 * mid));
    w[mid] = 1;
    for (int j = 1; j < mid; j++) w[mid + j] = 1ll * w[mid + j - 1] * wn % P;
  }
}
void ntt(int *f, int ty) {
  if (!ty) reverse(f + 1, f + len);
  for (int i = 0; i < len; i++)
    if (i < tr[i]) swap(f[i], f[tr[i]]);
  for (int mid = 1; mid < len; mid <<= 1)
    for (int i = 0; i < len; i += mid << 1)
      for (int j = 0; j < mid; j++) {
        int x = f[i + j], y = 1ll * f[i + j + mid] * w[mid + j] % P;
        m1(f[i + j] = x + y);
        m1(f[i + j + mid] = x - y + P);
      }
  if (ty) return;
  int invl = ksm(len, P - 2);
  for (int i = 0; i < len; i++) f[i] = 1ll * f[i] * invl % P;
}
void mul(const int *f, const int *g, int *h) {
  memcpy(A, f, len << 2);
  memcpy(B, g, len << 2);
  ntt(A, 1);
  ntt(B, 1);
  for (int i = 0; i < len; i++) A[i] = 1ll * A[i] * B[i] % P;
  ntt(A, 0);
  memcpy(h, A, len << 2);
}
}  // namespace poly
using namespace poly;
struct val {
  int st;
  int F1[N], F2[N], F3[N];
  val() {
    st = 0;
    memset(F1, 0, sizeof(F1));
    memset(F2, 0, sizeof(F2));
    memset(F3, 0, sizeof(F3));
  }
};
val v[31];
void calc(const int *f1, const int *f2, int *g) {
  for (int i = 1; i <= K; i++)
    m1(g[i] = f1[i - 1] + f2[i - 1]), m1(g[i] += f2[i]);
  g[0] = 1;
}
void Merge(const int *f1, const int *f2, const int *g1, const int *g2, int *h) {
  mul(f2, g2, h);
  mul(f1, g1, C);
  for (int i = 1; i <= K; i++) m1(h[i] += C[i - 1]);
  for (int i = K + 1; i < len; i++) h[i] = 0;
}
val operator*(const val &x, const val &y) {
  val res;
  res.st = x.st + y.st + 2;
  Merge(x.F1, x.F2, y.F1, y.F2, res.F1);
  Merge(x.F2, x.F3, y.F1, y.F2, res.F2);
  calc(res.F1, res.F2, res.F3);
  return res;
}
int main() {
  scanf("%d%d", &n, &K);
  pret(K + 1 << 1);
  prew(K + 1 << 1);
  v[0].st = 1;
  v[0].F1[0] = v[0].F1[1] = 1;
  v[0].F2[0] = 1;
  v[0].F2[1] = 3;
  v[0].F2[2] = 1;
  v[0].F3[0] = 1;
  v[0].F3[1] = v[0].F3[2] = 5;
  v[0].F3[3] = 1;
  int mk = 0;
  while (v[mk].st <= n) v[mk + 1] = v[mk] * v[mk], mk++;
  val ans = v[0];
  for (int i = mk; i >= 0; i--)
    if (ans.st + v[i].st + 2 <= n) ans = ans * v[i];
  while (ans.st < n) {
    calc(ans.F2, ans.F3, ans.F1);
    swap(ans.F1, ans.F2);
    swap(ans.F2, ans.F3);
    ans.st++;
  }
  for (int i = 1; i <= K; i++) printf("%d ", ans.F1[i]);
  return 0;
}