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>
const int _G = 3, mod = 998244353, Maxn = 135000, inv2 = 499122177;
inline int read() {
  register int X = 0;
  register char ch = 0;
  while (ch < 48 || ch > 57) ch = getchar();
  while (ch >= 48 && ch <= 57) X = X * 10 + (ch ^ 48), ch = getchar();
  return X;
}
inline void print(int *f, int len) {
  for (int i = 0; i < len; i++) printf("%d ", f[i]);
  puts("");
}
long long powM(long long a, long long t = mod - 2) {
  long long ans = 1;
  while (t) {
    if (t & 1) ans = ans * a % mod;
    a = a * a % mod;
    t >>= 1;
  }
  return ans;
}
const int invG = powM(_G);
int tr[Maxn << 1], tf;
void tpre(int n) {
  if (tf == n) return;
  tf = n;
  for (int i = 0; i < n; i++)
    tr[i] = (tr[i >> 1] >> 1) | ((i & 1) ? n >> 1 : 0);
}
void NTT(int *g, bool op, int n) {
  static unsigned long long f[Maxn << 1], w[Maxn] = {1};
  tpre(n);
  for (int i = 0; i < n; i++) f[i] = (((long long)mod << 5) + g[tr[i]]) % mod;
  for (int l = 1; l < n; l <<= 1) {
    unsigned long long tG = powM(op ? _G : invG, (mod - 1) / (l + l));
    for (int i = 1; i < l; i++) w[i] = w[i - 1] * tG % mod;
    for (int k = 0; k < n; k += l + l)
      for (int p = 0; p < l; p++) {
        int tt = w[p] * f[k | l | p] % mod;
        f[k | l | p] = f[k | p] + mod - tt;
        f[k | p] += tt;
      }
    if (l == (1 << 10))
      for (int i = 0; i < n; i++) f[i] %= mod;
  }
  if (!op) {
    unsigned long long invn = powM(n);
    for (int i = 0; i < n; ++i) g[i] = f[i] % mod * invn % mod;
  } else
    for (int i = 0; i < n; ++i) g[i] = f[i] % mod;
}
void px(int *f, int *g, int n) {
  for (int i = 0; i < n; ++i) f[i] = 1ll * f[i] * g[i] % mod;
}
int _g1[Maxn << 1];
void times(int *f, int *g, int len, int lim) {
  int n = 1;
  for (n; n < len + len; n <<= 1)
    ;
  memcpy(_g1, g, sizeof(int) * (n));
  for (int i = len; i < n; i++) _g1[i] = 0;
  NTT(f, 1, n);
  NTT(_g1, 1, n);
  px(f, _g1, n);
  NTT(f, 0, n);
  for (int i = lim; i < n; ++i) f[i] = 0;
  memset(_g1, 0, sizeof(int) * (n));
}
int _w2[Maxn << 1], _r2[Maxn << 1];
void invp(int *f, int m) {
  int n;
  for (n = 1; n < m; n <<= 1)
    ;
  _w2[0] = powM(f[0]);
  for (int len = 2; len <= n; len <<= 1) {
    for (int i = 0; i < (len >> 1); i++) _r2[i] = (_w2[i] << 1) % mod;
    memcpy(_g1, f, sizeof(int) * (len));
    NTT(_w2, 1, len << 1);
    px(_w2, _w2, len << 1);
    NTT(_g1, 1, len << 1);
    px(_w2, _g1, len << 1);
    NTT(_w2, 0, len << 1);
    memset(_w2 + len, 0, sizeof(int) * (len));
    for (int i = 0; i < len; i++) _w2[i] = (_r2[i] - _w2[i] + mod) % mod;
  }
  memcpy(f, _w2, sizeof(int) * (m));
  memset(_g1, 0, sizeof(int) * (n + n));
  memset(_w2, 0, sizeof(int) * (n + n));
  memset(_r2, 0, sizeof(int) * (n + n));
}
void dao(int *f, int m) {
  for (int i = 1; i < m; i++) f[i - 1] = 1ll * f[i] * i % mod;
  f[m - 1] = 0;
}
int inv[Maxn];
void Init(int lim) {
  inv[1] = 1;
  for (int i = 2; i <= lim; i++)
    inv[i] = 1ll * inv[mod % i] * (mod - mod / i) % mod;
}
void jifen(int *f, int m) {
  for (int i = m; i; i--) f[i] = 1ll * f[i - 1] * inv[i] % mod;
  f[0] = 0;
}
int _s3[Maxn << 2];
void lnp(int *f, int m) {
  memcpy(_s3, f, sizeof(int) * (m));
  invp(_s3, m);
  dao(f, m);
  times(f, _s3, m, m);
  jifen(f, m - 1);
  memset(_s3, 0, sizeof(int) * (m));
}
int _xp[Maxn << 1], _xp2[Maxn << 1];
void exp(int *f, int m) {
  int n = 1;
  for (; n < m; n <<= 1)
    ;
  _xp2[0] = 1;
  for (int len = 2; len <= n; len <<= 1) {
    memcpy(_xp, _xp2, sizeof(int) * (len >> 1));
    lnp(_xp, len);
    for (int i = 0; i < len; i++) _xp[i] = (f[i] - _xp[i] + mod) % mod;
    _xp[0] = (_xp[0] + 1) % mod;
    times(_xp2, _xp, len, len);
  }
  memcpy(f, _xp2, sizeof(int) * (m));
  memset(_xp, 0, sizeof(int) * (n + n));
  memset(_xp2, 0, sizeof(int) * (n + n));
}
void powp(int *f, int k, int n, int p) {
  static int g[Maxn] = {1};
  for (int m = 0; m < n - 1; m++) {
    g[m + 1] = 0;
    for (int i = 1; i < k; i++)
      g[m + 1] = (g[m + 1] +
                  (1ll * p * i - m + i - 1) % mod * f[i] % mod * g[m - i + 1]) %
                 mod;
    g[m + 1] = (long long)(g[m + 1] + mod) * inv[m + 1] % mod;
  }
  memcpy(f, g, sizeof(int) * (n));
}
int n, k, t, F1[Maxn << 1], F2[Maxn << 1];
int main() {
  scanf("%d%d", &n, &k);
  t = n < k ? n + 1 : k + 1;
  Init(t + 1);
  F1[0] = F2[0] = 1;
  F1[1] = F2[1] = 6;
  F1[2] = F2[2] = 1;
  powp(F1, 3, t, mod - inv2);
  powp(F2, 3, t, inv2);
  F2[0]++;
  F2[1]++;
  for (int i = 0; i < t; i++) F2[i] = 1ll * F2[i] * inv2 % mod;
  lnp(F2, t);
  for (int i = 0; i < t; i++) F2[i] = 1ll * F2[i] * (n + 1) % mod;
  exp(F2, t);
  times(F1, F2, t, t);
  for (int i = 1; i < t; i++) printf("%d ", F1[i]);
  for (int i = 0; i <= k - t; i++) printf("0 ");
  return 0;
}