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;
template <class T, class S>
ostream& operator<<(ostream& os, const pair<T, S>& v) {
  return os << "(" << v.first << ", " << v.second << ")";
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (int i = int(0); i <= int((static_cast<int>((v).size())) - 1); ++i) {
    if (i) os << ", ";
    os << v[i];
  }
  return os << "]";
}
template <class T>
bool setmax(T& _a, T _b) {
  if (_a < _b) {
    _a = _b;
    return true;
  }
  return false;
}
template <class T>
bool setmin(T& _a, T _b) {
  if (_b < _a) {
    _a = _b;
    return true;
  }
  return false;
}
template <class T>
T gcd(T _a, T _b) {
  return _b == 0 ? _a : gcd(_b, _a % _b);
}
long long powmod(long long a, long long b, long long m) {
  long long r = 1;
  for (; b > 0; b >>= 1, a = a * a % m) {
    if (b & 1) r = r * a % m;
  }
  return r;
}
const int MOD = 998244353;
const int GN = MOD - 1;
const int G = 3;
void fft(int* a, int n, bool is_inv) {
  static vector<int> b;
  if (static_cast<int>((b).size()) < n) b.resize(n);
  for (int t = n / 2; t >= 1; t /= 2) {
    long long w = powmod(G, GN / n * t, MOD), p = 1;
    if (is_inv) w = powmod(w, MOD - 2, MOD);
    for (int j = 0; j < n / 2; j += t, (p *= w) %= MOD)
      for (int i = 0; i < t; ++i) {
        long long u = a[i + j * 2];
        long long v = p * a[i + t + j * 2] % MOD;
        b[i + j] = int(u + v) % MOD;
        b[i + j + n / 2] = int(u + MOD - v) % MOD;
      }
    memcpy(a, &b[0], sizeof(int) * n);
  }
  if (is_inv) {
    long long inv = powmod(n, MOD - 2, MOD);
    for (int i = int(0); i <= int((n)-1); ++i) a[i] = int(a[i] * inv % MOD);
  }
}
struct Poly {
  vector<int> a;
  Poly operator*(const Poly& b) const {
    assert(static_cast<int>((a).size()) == static_cast<int>((b.a).size()));
    Poly r;
    r.a.resize(static_cast<int>((a).size()));
    for (int i = int(0); i <= int((static_cast<int>((a).size())) - 1); ++i)
      r.a[i] = ((long long)a[i] * b.a[i]) % MOD;
    return r;
  }
  void operator+=(const Poly& b) {
    if (static_cast<int>((a).size()) < static_cast<int>((b.a).size()))
      a.resize(static_cast<int>((b.a).size()));
    for (int i = int(0); i <= int((static_cast<int>((b.a).size())) - 1); ++i)
      (a[i] += b.a[i]) %= MOD;
  }
};
struct Mat {
  Poly a[2][2];
  Mat operator*(const Mat& b) const {
    Mat r;
    for (int i = int(0); i <= int((2) - 1); ++i)
      for (int j = int(0); j <= int((2) - 1); ++j)
        for (int k = int(0); k <= int((2) - 1); ++k) {
          r.a[i][j] += a[i][k] * b.a[k][j];
        }
    r.transform(true);
    for (int i = int(0); i <= int((2) - 1); ++i)
      for (int j = int(0); j <= int((2) - 1); ++j) {
        fill(r.a[i][j].a.begin() + static_cast<int>((r.a[i][j].a).size()) / 2,
             r.a[i][j].a.end(), 0);
      }
    r.transform(false);
    return r;
  }
  void transform(bool is_inv) {
    for (int i = int(0); i <= int((2) - 1); ++i)
      for (int j = int(0); j <= int((2) - 1); ++j)
        fft(&a[i][j].a[0], static_cast<int>((a[i][j].a).size()), is_inv);
  }
};
int main() {
  int n, k;
  cin >> n >> k;
  int len = 2;
  while (len < k * 2) len *= 2;
  Mat m;
  for (int i = int(0); i <= int((2) - 1); ++i)
    for (int j = int(0); j <= int((2) - 1); ++j) m.a[i][j].a.resize(len);
  m.a[0][0].a[0] = m.a[0][0].a[1] = 1;
  m.a[0][1].a[1] = 1;
  m.a[1][0].a[0] = 1;
  Mat r;
  for (int i = int(0); i <= int((2) - 1); ++i)
    for (int j = int(0); j <= int((2) - 1); ++j) r.a[i][j].a.resize(len);
  r.a[0][0].a[0] = 1;
  r.a[1][1].a[0] = 1;
  m.transform(false);
  r.transform(false);
  for (; n > 0; n >>= 1, m = m * m) {
    if (n & 1) r = r * m;
  }
  r.transform(true);
  for (int i = int(1); i <= int(k); ++i) {
    long long cur = 0;
    cur += r.a[1][0].a[i] + r.a[1][0].a[i - 1];
    cur += r.a[1][1].a[i];
    printf("%d ", int(cur % MOD));
  }
  printf("\n");
  return 0;
}
