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 long long N = 262150;
const long long mod = 998244353;
long long qpow(long long p, long long q) {
  return (q & 1 ? p : 1) * (q ? qpow(p * p % mod, q / 2) : 1) % mod;
}
namespace cipolla {
inline long long le(long long x) { return qpow(x, (mod - 1) / 2); }
long long w;
struct comp {
  long long x, y;
  comp(long long a = 0, long long b = 0) {
    x = a;
    y = b;
  }
};
comp operator+(comp a, comp b) {
  return comp((a.x + b.x) % mod, (a.y + b.y) % mod);
}
comp operator-(comp a, comp b) {
  return comp((a.x - b.x + mod) % mod, (a.y - b.y + mod) % mod);
}
comp operator*(comp a, comp b) {
  return comp((a.x * b.x + a.y * b.y % mod * w) % mod,
              (a.x * b.y + a.y * b.x) % mod);
}
comp operator^(comp a, long long b) {
  comp o(1, 0);
  for (; b; a = a * a, b >>= 1)
    if (b & 1) o = o * a;
  return o;
}
long long calc(long long x) {
  x %= mod;
  long long a;
  while (true) {
    a = rand();
    w = (a * a - x + mod) % mod;
    if (le(w) == mod - 1) break;
  }
  comp s = comp(a, 1) ^ ((mod + 1) / 2);
  return min(s.x, mod - s.x);
}
}  // namespace cipolla
namespace po {
long long rev[N], inv[N], w[N], sz;
void presolve(long long l) {
  long long len = 1;
  sz = 0;
  while (len < l) len <<= 1, ++sz;
  for (long long i = 1; i < len; i++) {
    inv[i] = (i == 1 ? 1 : inv[mod % i] * (mod - mod / i) % mod);
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (sz - 1));
  }
  long long wn = qpow(3, (mod - 1) / len);
  w[len / 2] = 1;
  for (long long i = len / 2 + 1; i < len; i++) w[i] = w[i - 1] * wn % mod;
  for (long long i = len / 2 - 1; i; i--) w[i] = w[i << 1];
}
long long pre(long long l) {
  long long g;
  for (g = 1; g < l; g <<= 1)
    ;
  return g;
}
void ntt(long long *a, long long o, long long n) {
  static unsigned long long s[N];
  long long t = sz - __builtin_ctz(n), x;
  for (long long i = 0; i < n; i++) s[rev[i] >> t] = a[i];
  for (long long l = 1; l < n; l <<= 1)
    for (long long i = 0; i < n; i += l << 1)
      for (long long j = 0; j < l; j++) {
        x = s[i + j + l] * w[j + l] % mod;
        s[i + j + l] = s[i + j] + mod - x;
        s[i + j] += x;
      }
  for (long long i = 0; i < n; i++) a[i] = s[i] % mod;
  if (o) {
    x = qpow(n, mod - 2);
    for (long long i = 0; i < n; i++) a[i] = a[i] * x % mod;
    reverse(a + 1, a + n);
  }
}
void mult(long long n, long long *x, long long *y, long long *z) {
  static long long a[N], b[N];
  long long l = pre(n << 1);
  for (long long i = 0; i < l; i++) {
    a[i] = (i < n ? x[i] : 0);
    b[i] = (i < n ? y[i] : 0);
  }
  ntt(a, 0, l);
  ntt(b, 0, l);
  for (long long i = 0; i < l; i++) z[i] = a[i] * b[i] % mod;
  ntt(z, 1, l);
  for (long long i = n; i < l; i++) z[i] = 0;
}
void inve(long long len, long long *a, long long *b) {
  if (len == 1)
    *b = qpow(*a, mod - 2);
  else {
    inve((len + 1) / 2, a, b);
    static long long c[N];
    long long n = pre(len << 1);
    for (long long i = 0; i < n; i++) i < len ? c[i] = a[i] : b[i] = c[i] = 0;
    ntt(b, 0, n);
    ntt(c, 0, n);
    for (long long i = 0; i < n; i++)
      b[i] = ((b[i] + b[i] - b[i] * b[i] % mod * c[i]) % mod + mod) % mod;
    ntt(b, 1, n);
    for (long long i = len; i < n; i++) b[i] = 0;
  }
}
void sqrt(long long n, long long *a, long long *b) {
  if (n == 1)
    *b = cipolla::calc(*a);
  else {
    sqrt((n + 1) / 2, a, b);
    static long long c[N];
    inve(n, b, c);
    mult(n, a, c, c);
    for (long long i = 0; i < n; i++) b[i] = (b[i] + c[i]) * inv[2] % mod;
  }
}
void deri(long long n, long long *a, long long *b) {
  for (long long i = 0; i < n - 1; i++) b[i] = a[i + 1] * (i + 1) % mod;
  b[n - 1] = 0;
}
void inte(long long n, long long *a, long long *b) {
  for (long long i = n - 1; i > 0; --i) b[i] = a[i - 1] * inv[i] % mod;
  b[0] = 0;
}
void loge(long long n, long long *a, long long *b) {
  static long long c[N];
  inve(n, a, b);
  deri(n, a, c);
  mult(n, b, c, b);
  inte(n, b, b);
}
void expr(long long n, long long *a, long long *b) {
  if (n == 1)
    *b = 1;
  else {
    expr((n + 1) / 2, a, b);
    static long long c[N];
    loge(n, b, c);
    for (long long i = 0; i < n; i++) c[i] = (a[i] - c[i] + mod) % mod;
    c[0] = (c[0] + 1) % mod;
    mult(n, b, c, b);
  }
}
}  // namespace po
long long n, k, a[N], b[N], c[N];
struct poly {
  vector<long long> x;
  poly operator+(const poly &o) {
    poly res;
    for (long long i = 0; i < k && i < x.size() && i < o.x.size(); i++) {
      res.x.push_back((x[i] + o.x[i]) % mod);
    }
    return res;
  }
  poly operator*(const poly &o) {
    poly res;
    for (long long i = 0; i < k; i++) a[i] = b[i] = c[i] = 0;
    for (long long i = 0; i < x.size(); i++) a[i] = x[i];
    for (long long i = 0; i < o.x.size(); i++) b[i] = o.x[i];
    po::mult(k, a, b, c);
    for (long long i = 0; i < k; i++) res.x.push_back(c[i]);
    return res;
  }
  poly shift() {
    poly res;
    res.x.push_back(0);
    for (long long i = 1; i < k; i++) res.x.push_back(x[i - 1]);
    return res;
  }
};
poly nullpoly() {
  poly res;
  for (long long i = 0; i < k; i++) res.x.push_back(0);
  return res;
}
poly unitpoly() {
  poly res;
  res.x.push_back(1);
  for (long long i = 1; i < k; i++) res.x.push_back(0);
  return res;
}
void print(const vector<long long> &vec) {
  for (long long i = 1; i < min(1ull * k, 1ull * vec.size()); i++)
    cout << vec[i] << " ";
  cout << endl;
}
void print(const poly &p) { print(p.x); }
signed main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  ++k;
  po::presolve((k + 1) << 2);
  poly f2, f1, f0;
  f0 = unitpoly();
  f1 = f0 + f0.shift();
  f2 = f1 + f1.shift() + f0.shift();
  if (n <= 2) {
    if (n == 1) print(f1);
    if (n == 2) print(f2);
  } else {
    swap(f2, f0);
    long long tmp = n;
    vector<long long> vop;
    while (tmp > 2) {
      if (tmp & 1) {
        vop.push_back(1);
        tmp--;
      } else {
        vop.push_back(0);
        tmp >>= 1;
      }
    }
    reverse(vop.begin(), vop.end());
    for (long long op : vop) {
      if (op == 0) {
        poly g2, g1, g0;
        g0 = f0 * f0 + (f1 * f1).shift();
        g1 = f0 * f1 + (f1 * f2).shift();
        g2 = f1 * f1 + (f2 * f2).shift();
        f0 = g0;
        f1 = g1;
        f2 = g2;
      } else {
        poly tmp;
        tmp = f0 + f0.shift() + f1.shift();
        f2 = f1;
        f1 = f0;
        f0 = tmp;
      }
    }
    print(f0);
  }
  return 0;
}