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;
long long n, m, k;
const long long mod = 998244353;
const long long G = 3;
const long long maxn = 300001;
long long read() {
  long long s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * f;
}
long long read2() {
  long long x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) x = (1ll * x * 10 + ch - '0') % mod, ch = getchar();
  return x;
}
inline void print(long long *f, long long len) {
  for (long long i = 0; i < len; i++) printf("%lld ", f[i]);
  puts("");
}
long long qpow(long long a, long long b = mod - 2) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
const long long invG = qpow(G);
long long tr[maxn], f[7][maxn], g[7][maxn], invn, trs;
long long inv[maxn];
void poly_pre(long long n) {
  if (trs == n) return;
  trs = n;
  for (long long i = 0; i < n; ++i)
    tr[i] = (tr[i >> 1] >> 1) | ((i & 1) ? n >> 1 : 0);
}
void ntt(long long *f, long long n, bool op) {
  poly_pre(n);
  for (long long i = 0; i < n; ++i) {
    if (i < tr[i]) swap(f[i], f[tr[i]]);
  }
  for (long long p = 2; p <= n; p <<= 1) {
    long long len = (p >> 1);
    long long bas = qpow(op == 1 ? G : invG, (mod - 1) / p);
    for (long long k = 0; k < n; k += p) {
      long long buf = 1;
      for (long long l = k; l < k + len; ++l) {
        long long kk = buf * f[l + len] % mod;
        f[l + len] = f[l] - kk;
        if (f[len + l] < 0) f[len + l] += mod;
        f[l] = f[l] + kk;
        f[l] %= mod;
        buf *= bas, buf %= mod;
      }
    }
  }
  if (!op) {
    long long invn = qpow(n);
    for (long long i = 0; i < n; ++i) f[i] = (f[i] * invn) % mod;
  }
}
void poly_mul(long long *f, long long *g, long long n) {
  for (long long i = 0; i <= n; ++i) f[i] = f[i] * g[i] % mod;
}
void times(long long *f, long long *g, long long len, long long lim) {
  static long long sav[maxn << 1];
  long long n;
  for (n = 1; n < len + len; n <<= 1)
    ;
  memset(sav, 0, sizeof(long long) * (n)),
      memcpy(sav, g, sizeof(long long) * (n));
  ntt(f, n, 1), ntt(sav, n, 1);
  poly_mul(f, sav, n);
  ntt(f, n, 0);
  memset(f + lim, 0, sizeof(long long) * (n - lim)),
      memset(sav, 0, sizeof(long long) * (n));
}
long long w[maxn], r[maxn], sav[maxn];
void poly_inv(long long *f, long long m) {
  long long n;
  for (n = 1; n < m; n <<= 1)
    ;
  w[0] = qpow(f[0]);
  for (long long len = 2; len <= n; len <<= 1) {
    for (long long i = 0; i < (len >> 1); ++i) {
      r[i] = (w[i] << 1) % mod;
    }
    memcpy(sav, f, sizeof(long long) * (len));
    ntt(w, len << 1, 1);
    ntt(sav, len << 1, 1);
    poly_mul(w, w, len << 1), poly_mul(w, sav, len << 1);
    ntt(w, len << 1, 0);
    for (long long i = 0; i < len; i++) w[i] = (r[i] - w[i] + mod) % mod;
    memset(w + len, 0, sizeof(long long) * (len));
  }
  memcpy(f, w, sizeof(long long) * (m));
  memset(sav, 0, sizeof(long long) * (n << 1)),
      memset(w, 0, sizeof(long long) * (n << 1)),
      memset(r, 0, sizeof(long long) * (n << 1));
  return;
}
long long b1[maxn << 1], b2[maxn << 1];
void poly_sqrt(long long *f, long long m) {
  long long n;
  for (n = 1; n <= m; n <<= 1)
    ;
  b1[0] = 1;
  for (long long len = 2; len <= n; len <<= 1) {
    for (long long i = 0; i < (len >> 1); ++i) {
      b2[i] = (b1[i] << 1) % mod;
    }
    poly_inv(b2, len);
    ntt(b1, len, 1);
    poly_mul(b1, b1, len);
    ntt(b1, len, 0);
    for (long long i = 0; i < len; ++i) {
      b1[i] = (b1[i] + f[i]) % mod;
    }
    times(b1, b2, len, len);
  }
  memcpy(f, b1, sizeof(long long) * (m));
  memset(b1, 0, sizeof(long long) * (n << 1)),
      memset(b2, 0, sizeof(long long) * (n << 1));
}
void poly_dao(long long *f, long long m) {
  for (long long i = 1; i < m; ++i) {
    f[i - 1] = f[i] * 1ll * i % mod;
  }
  f[m - 1] = 0;
}
void initmath(long long lim) {
  inv[1] = 1;
  for (long long i = 2; i <= lim; i++)
    inv[i] = 1ll * inv[mod % i] * (mod - mod / i) % mod;
}
void poly_jifen(long long *f, long long m) {
  for (long long i = m; i; --i) {
    f[i] = 1ll * f[i - 1] * inv[i] % mod;
  }
  f[0] = 0;
}
void poly_ln(long long *f, long long m) {
  static long long g[maxn << 1];
  memcpy(g, f, sizeof(long long) * (m));
  poly_inv(g, m);
  poly_dao(f, m);
  times(f, g, m, m);
  poly_jifen(f, m - 1);
  memset(g, 0, sizeof(long long) * (m));
}
void poly_exp(long long *f, long long m) {
  static long long s1[maxn], s2[maxn];
  long long n = 1;
  for (n = 1; n <= m; n <<= 1)
    ;
  s2[0] = 1;
  for (long long len = 2; len <= n; len <<= 1) {
    memcpy(s1, s2, sizeof(long long) * (len >> 1));
    poly_ln(s1, len);
    for (long long i = 0; i < len; ++i) {
      s1[i] = (f[i] - s1[i] + mod) % mod;
    }
    s1[0]++;
    s1[0] %= mod;
    times(s2, s1, len, len);
  }
  memcpy(f, s2, sizeof(long long) * (m));
  memset(s1, 0, sizeof(long long) * (n)),
      memset(s2, 0, sizeof(long long) * (n));
}
void poly_pow(long long *f, long long m, long long k) {
  poly_ln(f, m);
  for (long long i = 0; i < m; ++i) f[i] = (f[i] * k) % mod;
  poly_exp(f, m);
}
void poly_div(long long *f, long long *g, long long n, long long m) {
  static long long q[maxn], t[maxn];
  long long len = n - m + 1;
  reverse(g, g + m);
  memcpy(q, g, sizeof(long long) * (len));
  reverse(g, g + m);
  reverse(f, f + n);
  memcpy(t, f, sizeof(long long) * (len));
  reverse(f, f + n);
  poly_inv(q, len);
  times(q, t, len, len);
  reverse(q, q + len);
  times(g, q, n, n);
  for (long long i = 0; i < m - 1; ++i) g[i] = (f[i] - g[i] + mod) % mod;
  memset(g + m + 1, 0, sizeof(long long) * (len));
  memcpy(f, q, sizeof(long long) * (len));
  memset(f + len, 0, sizeof(long long) * (n - len));
}
void solve(long long nw) {
  long long cnt = 1, lim = 0;
  while (cnt <= nw * 2) {
    cnt <<= 1;
    lim++;
  }
  lim--;
  for (long long i = 0, iend = cnt - 1; i <= iend; ++i) {
    g[0][i] = g[1][i] = g[2][i] = g[3][i] = g[4][i] = 0;
  }
  ntt(f[0], cnt, 1);
  ntt(f[1], cnt, 1);
  ntt(f[2], cnt, 1);
  for (long long i = 0, iend = cnt - 1; i <= iend; ++i) {
    g[0][i] = f[0][i] * f[0][i] % mod;
    g[1][i] = f[1][i] * f[1][i] % mod;
    g[2][i] = f[2][i] * f[2][i] % mod;
    g[3][i] = f[0][i] * f[1][i] % mod;
    g[4][i] = f[1][i] * f[2][i] % mod;
    f[0][i] = 0;
    f[1][i] = 0;
    f[2][i] = 0;
  }
  for (long long i = 0, iend = 4; i <= iend; ++i) ntt(g[i], cnt, 0);
  for (long long i = 0, iend = cnt - 1; i <= iend; ++i) {
    f[0][i] = g[0][i];
    if (i) f[0][i] += g[1][i - 1];
    f[1][i] = g[3][i];
    if (i) f[1][i] += g[4][i - 1];
    f[2][i] = g[1][i];
    if (i) f[2][i] += g[2][i - 1];
    f[0][i] %= mod;
    f[1][i] %= mod;
    f[2][i] %= mod;
  }
  for (long long i = 0, iend = cnt - 1; i <= iend; ++i) {
    g[0][i] = g[1][i] = g[2][i] = g[3][4] = g[4][i] = 0;
  }
  for (long long i = nw, iend = cnt - 1; i <= iend; ++i)
    f[0][i] = f[1][i] = f[2][i] = 0;
  return;
}
void change(long long nw) {
  for (long long i = 0, iend = nw - 1; i <= iend; ++i) {
    f[2][i] = f[1][i];
    f[1][i] = f[0][i];
    f[0][i] = 0;
  }
  f[0][0] = 1;
  for (long long i = 1, iend = nw - 1; i <= iend; ++i) {
    f[0][i] = (f[1][i - 1] + f[2][i - 1] + f[1][i]) % mod;
  }
  return;
}
signed main() {
  n = read(), k = read();
  f[0][0] = 1;
  f[1][0] = f[0][1] = 1;
  long long lim = log2(n);
  for (long long i = lim - 1, iend = 0; i >= iend; --i) {
    solve(k + 10);
    if (n & (1 << i)) {
      change(k + 10);
    }
  }
  for (long long i = 1, iend = k; i <= iend; ++i) printf("%lld ", f[0][i]);
}