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;
struct IO_Tp {
  const static int _I_Buffer_Size = 2 << 22;
  char _I_Buffer[_I_Buffer_Size], *_I_pos = _I_Buffer;
  const static int _O_Buffer_Size = 2 << 22;
  char _O_Buffer[_O_Buffer_Size], *_O_pos = _O_Buffer;
  IO_Tp() { fread(_I_Buffer, 1, _I_Buffer_Size, stdin); }
  ~IO_Tp() { fwrite(_O_Buffer, 1, _O_pos - _O_Buffer, stdout); }
  IO_Tp &operator>>(long long &res) {
    long long f = 1;
    while (!isdigit(*_I_pos) && (*_I_pos) != '-') ++_I_pos;
    if (*_I_pos == '-') f = -1, ++_I_pos;
    res = *_I_pos++ - '0';
    while (isdigit(*_I_pos)) res = res * 10 + (*_I_pos++ - '0');
    res *= f;
    return *this;
  }
  IO_Tp &operator<<(long long n) {
    if (n < 0) *_O_pos++ = '-', n = -n;
    static char _buf[10];
    char *_pos = _buf;
    do *_pos++ = '0' + n % 10;
    while (n /= 10);
    while (_pos != _buf) *_O_pos++ = *--_pos;
    return *this;
  }
  IO_Tp &operator<<(char ch) {
    *_O_pos++ = ch;
    return *this;
  }
} IO;
const long long N = 1048576, P = 998244353;
long long inv[N];
namespace Poly {
const long long G = 3, img = 86583718;
long long lmt, rev[N], a[N], b[N], c[N], d[N], e[N], h[N], x[N], y[N], z[N],
    X[N], Y[N], ff[N], gg[N], iv[N], t[N];
inline long long qpow(long long a, long long k) {
  long long ret = 1;
  while (k) {
    if (k & 1) ret = ret * a % P;
    a = (a * a) % P;
    k >>= 1;
  }
  return ret % P;
}
inline void init(long long n) {
  lmt = 1;
  long long t = 0;
  while (lmt < n) lmt <<= 1, t++;
  for (long long i = 1; i < lmt; i++)
    rev[i] = (rev[i >> 1] >> 1) | (i & 1) << (t - 1);
}
inline void NTT(long long *A, long long lmt, long long tp) {
  for (long long i = 0; i < lmt; i++)
    if (i < rev[i]) swap(A[i], A[rev[i]]);
  for (long long m = 1; m < lmt; m <<= 1)
    for (long long j = 0, Wn = qpow(G, (P - 1) / (m << 1)); j < lmt;
         j += m << 1)
      for (long long k = 0, w = 1, x, y; k < m; k++, w = w * Wn % P)
        x = A[j + k], y = w * A[j + k + m] % P, A[j + k] = (x + y) % P,
        A[j + k + m] = (x - y + P) % P;
  if (tp == 1) return;
  reverse(A + 1, A + lmt);
  for (long long i = 0, inv = qpow(lmt, P - 2); i <= lmt; i++)
    A[i] = A[i] * inv % P;
}
inline void mul(long long *f, long long *g, long long len) {
  init(len << 1);
  NTT(f, lmt, 1);
  NTT(g, lmt, 1);
  for (long long i = 0; i < lmt; i++) f[i] = (f[i] * g[i]) % P;
  NTT(f, lmt, -1);
}
void getinv(long long *f, long long *g, long long len) {
  if (len == 1) {
    g[0] = qpow(f[0], P - 2);
    return;
  }
  getinv(f, g, len + 1 >> 1);
  init(len << 1);
  for (long long i = 0; i < len; i++) c[i] = f[i];
  for (long long i = len; i < lmt; i++) c[i] = 0;
  NTT(c, lmt, 1), NTT(g, lmt, 1);
  for (long long i = 0; i < lmt; i++)
    g[i] = (2LL - g[i] * c[i] % P + P) % P * g[i] % P;
  NTT(g, lmt, -1);
  for (long long i = len; i < lmt; i++) g[i] = 0;
  memset((c), 0, len << 5);
}
inline void getdev(long long *f, long long *g, long long len) {
  for (long long i = 1; i < len; i++) g[i - 1] = i * f[i] % P;
  g[len - 1] = g[len] = 0;
}
inline void getinvdev(long long *f, long long *g, long long len) {
  for (long long i = 1; i <= len; i++) g[i] = f[i - 1] * inv[i] % P;
  g[0] = 0;
}
inline void getln(long long *f, long long *g, long long len) {
  getdev(f, a, len);
  getinv(f, b, len);
  mul(a, b, len);
  getinvdev(a, g, len);
  memset((a), 0, len << 5), memset((b), 0, len << 5);
}
void getexp(long long *f, long long *g, long long len) {
  if (len == 1) {
    g[0] = 1;
    return;
  }
  getexp(f, g, len + 1 >> 1);
  init(len << 1);
  getln(g, d, len);
  for (long long i = 0; i < len; i++) e[i] = f[i];
  NTT(g, lmt, 1), NTT(d, lmt, 1), NTT(e, lmt, 1);
  for (long long i = 0; i < lmt; i++) g[i] = (1 - d[i] + e[i] + P) * g[i] % P;
  NTT(g, lmt, -1);
  for (long long i = len; i < lmt; i++) g[i] = 0;
  memset((d), 0, len << 5), memset((e), 0, len << 5);
}
void getpow(long long *f, long long *g, long long len, long long k) {
  getln(f, h, len);
  for (long long i = 0; i < len; i++) h[i] = h[i] * k % P;
  getexp(h, g, len);
  memset((h), 0, len << 5);
}
}  // namespace Poly
using Poly::getpow;
using Poly::mul;
long long prep(long long n) {
  long long lmt = 1;
  inv[0] = inv[1] = 1;
  while (lmt < n) lmt <<= 1;
  for (long long i = 2; i < lmt; i++) inv[i] = (P - P / i) * inv[P % i] % P;
  return lmt;
}
long long n, t, k, f[N], g[N], h[N];
int main() {
  IO >> n >> t;
  k = min(n, t) + 1;
  prep(k << 1);
  f[0] = 1, f[1] = 3, g[0] = 1, g[1] = P - 3;
  for (long long i = 2; i < k; i++) {
    long long t = ((6 * i - 9) * f[i - 1] + (i - 3) * f[i - 2]) % P;
    f[i] = (P - t) * inv[i] % P;
    t = ((6 * i - 3) * g[i - 1] + (i - 1) * g[i - 2]) % P;
    g[i] = (P - t) * inv[i] % P;
  }
  f[1]++, f[0]++;
  for (long long i = 0; i < k; i++) f[i] = f[i] * inv[2] % P;
  getpow(f, h, k, n + 1);
  mul(g, h, k);
  for (long long i = 1; i < k; i++) IO << g[i] << ' ';
  for (long long i = k; i <= t; i++) IO << 0LL << ' ';
  return 0;
}
