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;
namespace io {
const int L = (1 << 19) + 1;
char ibuf[L], *iS, *iT, obuf[L], *oS = obuf, *oT = obuf + L - 1, c, st[55];
int f, tp;
inline char gc() {
  if (iS == iT) {
    iT = (iS = ibuf) + fread(ibuf, 1, L, stdin);
    return iS == iT ? EOF : *iS++;
  }
  return *iS++;
}
inline void flush() {
  fwrite(obuf, 1, oS - obuf, stdout);
  oS = obuf;
}
inline void putc(char x) {
  *oS++ = x;
  if (oS == oT) flush();
}
template <class I>
inline void gi(I &x) {
  for (f = 1, c = gc(); c < '0' || c > '9'; c = gc())
    if (c == '-') f = -1;
  for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15);
  x *= f;
}
template <class I>
inline void print(I x) {
  if (!x) putc('0');
  if (x < 0) putc('-'), x = -x;
  while (x) st[++tp] = x % 10 + '0', x /= 10;
  while (tp) putc(st[tp--]);
}
inline void gs(char *s, int &l) {
  for (c = gc(); c != '_' && (c < 'a' || c > 'z'); c = gc())
    ;
  for (l = 0; c == '_' || c <= 'z' && c >= 'a'; c = gc()) s[l++] = c;
}
inline void give() {
  putc('n');
  putc('o');
  putc('n');
  putc('e');
}
};  // namespace io
using io::gi;
using io::gs;
using io::print;
using io::putc;
const int N = 1050005, P = 998244353, G = 23, g = 3;
inline int fpow(int a, int t) {
  static int r;
  for (r = 1; t; t >>= 1, a = (long long)a * a % P)
    if (t & 1) r = (long long)r * a % P;
  return r;
}
int *_ww[G], *_iww[G], inv[N];
inline int nxp(int n) {
  int l = 1;
  while (l <= n) l <<= 1;
  return l;
}
void predo(int n) {
  static int poc[N << 2], *iter = poc;
  for (int l = 2, i = 1; l <= n; l <<= 1, i++) {
    _ww[i] = iter;
    for (int j = *iter = 1, k = l >> 1, w = fpow(g, (P - 1) / l); j < k; j++)
      iter[j] = (long long)iter[j - 1] * w % P;
    iter += l >> 1;
    _iww[i] = iter;
    for (int j = *iter = 1, k = l >> 1, w = fpow(g, (P - 1) / l * (l - 1));
         j < k; j++)
      iter[j] = (long long)iter[j - 1] * w % P;
    iter += l >> 1;
  }
  inv[1] = 1;
  for (int i = 2; i <= n; i++) inv[i] = P - (long long)(P / i) * inv[P % i] % P;
}
void fft(int a[], int n, int dft) {
  int i, j, k, l, u, v, t, *_w;
  for (i = 1, j = n >> 1; i < n - 1; i++) {
    if (i < j) swap(a[i], a[j]);
    for (k = n >> 1; (j ^= k) < k; k >>= 1)
      ;
  }
  for (l = 2, t = 1; l <= n; l <<= 1, t++) {
    if (dft == 1)
      _w = _ww[t];
    else
      _w = _iww[t];
    for (i = l >> 1, j = 0; j < n; j += l)
      for (k = 0; k < i; k++)
        u = a[j + k], v = (long long)a[j + k + i] * _w[k] % P,
        a[j + k] = (u + v) % P, a[j + k + i] = (u - v + P) % P;
  }
  if (dft == -1)
    for (i = 0; i < n; i++) a[i] = (long long)a[i] * inv[n] % P;
}
void convol(int a[], int na, int b[], int nb, int c[], int n) {
  static int a0[N], b0[N], c0[N];
  assert(na >= 0);
  assert(nb >= 0);
  assert(n >= 0);
  int m = nxp(na + nb - 2);
  memset(a0, 0, sizeof((a0)[0]) * (m));
  memset(b0, 0, sizeof((b0)[0]) * (m));
  memcpy(a0, a, sizeof((a)[0]) * (na));
  memcpy(b0, b, sizeof((b)[0]) * (nb));
  fft(a0, m, 1);
  fft(b0, m, 1);
  for (int i = 0; i < m; i++) c0[i] = (long long)a0[i] * b0[i] % P;
  fft(c0, m, -1);
  memcpy(c, c0, sizeof((c0)[0]) * (n));
}
void Inv(int a[], int b[], int n) {
  if (n == 1) return (void)(*b = fpow(*a, P - 2));
  static int w[N];
  int D = (n + 1) >> 1, m = nxp(n + D * 2 - 3);
  Inv(a, w, D);
  static int a0[N], b0[N], b1[N];
  memset(a0, 0, sizeof((a0)[0]) * (m));
  memset(b0, 0, sizeof((b0)[0]) * (m));
  memcpy(a0, a, sizeof((a)[0]) * (n));
  memcpy(b0, w, sizeof((w)[0]) * (D));
  fft(a0, m, 1);
  fft(b0, m, 1);
  for (int i = 0; i < m; i++)
    b1[i] = ((2 * b0[i] - (long long)a0[i] * b0[i] % P * b0[i]) % P + P) % P;
  fft(b1, m, -1);
  memcpy(b, b1, sizeof((b1)[0]) * (n));
}
void Ln(int a[], int b[], int n) {
  if (n == 1) return (void)(*b = 0);
  static int ia[N], da[N], b0[N];
  Inv(a, ia, n - 1);
  for (int i = 1; i < n; i++) da[i - 1] = (long long)a[i] * i % P;
  convol(ia, n - 1, da, n - 1, b0, n - 1);
  b[0] = 0;
  for (int i = 1; i < n; i++) b[i] = (long long)b0[i - 1] * inv[i] % P;
}
void Exp(int a[], int b[], int n) {
  if (n == 1) return (void)(*b = 1);
  static int b0[N], _b0[N], w0[N], w1[N], w2[N];
  int D = (n + 1) >> 1, m = nxp(n + D - 2);
  memset(b0, 0, sizeof((b0)[0]) * (n));
  Exp(a, b0, D);
  Ln(b0, _b0, n);
  memcpy(w0, b0, sizeof((b0)[0]) * (D));
  for (int i = 0; i < n; i++) w1[i] = (a[i] - _b0[i] + P) % P;
  w1[0]++;
  convol(w0, D, w1, n, w2, n);
  memcpy(b, w2, sizeof((w2)[0]) * (n));
}
int n, m;
int delta[N], lnDelta[N], sqrtDelta[N];
int lambda[N], lnLambda[N], lnAns[N], ans[N];
int main() {
  gi(n);
  gi(m);
  predo(nxp(m * 2));
  m++;
  delta[0] = 1;
  delta[1] = 6;
  delta[2] = 1;
  Ln(delta, lnDelta, m);
  for (int i = 0; i < m; i++)
    lnDelta[i] = (long long)lnDelta[i] * (P / 2 + 1) % P;
  Exp(lnDelta, sqrtDelta, m);
  lambda[0] = lambda[1] = 1;
  for (int i = 0; i < m; i++)
    lambda[i] = (long long)(lambda[i] + sqrtDelta[i]) * (P / 2 + 1) % P;
  Ln(lambda, lnLambda, m);
  for (int i = 0; i < m; i++)
    lnAns[i] = ((long long)lnLambda[i] * (n + 1) - lnDelta[i] + P) % P;
  Exp(lnAns, ans, m);
  for (int i = n + 1; i < m; i++) ans[i] = 0;
  for (int i = 1; i < m; i++) print(ans[i]), putc(' ');
  io::flush();
  return 0;
}
