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 int MOD = (int)1e9 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T1, class T2>
inline int chkmin(T1& a, const T2& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T1, class T2>
inline int chkmax(T1& a, const T2& val) {
  return a < val ? a = val, 1 : 0;
}
template <class T>
inline T isqrt(T k) {
  T r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
template <class T>
inline T icbrt(T k) {
  T r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return (int)fpow(a, p - 2, p); }
const int maxn = 1 << 20;
const int mod = 998244353;
void fft(int a[], int p, int r, int n) {
  for (int m = n, h; h = m / 2, m >= 2; r = (long long)r * r % p, m = h) {
    for (int i = 0, w = 1; i < h; i++, w = (long long)w * r % p) {
      for (int j = i; j < n; j += m) {
        int k = j + h, x = a[j] - a[k];
        if (x < 0) x += p;
        a[j] += a[k];
        if (a[j] >= p) a[j] -= p;
        a[k] = (long long)w * x % p;
      }
    }
  }
  for (int i = 0, j = 1; j < n - 1; j++) {
    for (int k = n / 2; k > (i ^= k); k /= 2)
      ;
    if (j < i) swap(a[i], a[j]);
  }
}
int n, k;
int c[maxn];
int fa[maxn];
int fb[maxn];
void multiply(vector<int>& a, vector<int>& b, int c[], int p) {
  int na = int((a).size()), nb = int((b).size());
  int n = 1, d = p - 1;
  while (n <= na + nb) n <<= 1, d >>= 1;
  fill_n(fa, n, 0), fill_n(fb, n, 0);
  for (int i = (0); i < (na); i++) fa[i] = a[i];
  for (int i = (0); i < (nb); i++) fb[i] = b[i];
  int w1 = fpow(3, d, p);
  int w2 = inv(w1, p);
  fft(fa, p, w1, n);
  fft(fb, p, w1, n);
  for (int i = (0); i < (n); i++) fa[i] = mult(fa[i], fb[i], p);
  fft(fa, p, w2, n);
  int rem = inv(n, p);
  for (int i = (0); i < (n); i++) c[i] = mult(fa[i], rem, p);
}
struct Num {
  vector<int> a, b;
  Num() {}
  Num(vector<int> a, vector<int> b) : a(a), b(b) {}
  Num(const Num& rhs) : a(rhs.a), b(rhs.b) {}
};
Num merge(Num& x, Num& y) {
  Num res;
  int ka = max(int((x.a).size()) + int((y.a).size()) - 1,
               int((x.b).size()) + int((y.b).size()) + 1);
  ka = min(ka, k + 1);
  int kb = max(int((x.a).size()) + int((y.b).size()) - 1,
               int((x.b).size()) + int((y.a).size()) - 1);
  kb = min(kb, k + 1);
  res.a.resize(ka, 0);
  res.b.resize(kb, 0);
  multiply(x.a, y.a, c, mod);
  for (int i = (0); i < (ka); i++) addmod(res.a[i], c[i], mod);
  multiply(x.b, y.b, c, mod);
  static int cof[] = {1, 6, 1};
  for (int i = (0); i < (ka); i++)
    for (int j = (0); j < (3); j++) {
      if (i + j < ka) {
        addmod(res.a[i + j], mult(c[i], cof[j], mod), mod);
      }
    }
  multiply(x.a, y.b, c, mod);
  for (int i = (0); i < (kb); i++) {
    addmod(res.b[i], c[i], mod);
  }
  multiply(x.b, y.a, c, mod);
  for (int i = (0); i < (kb); i++) {
    addmod(res.b[i], c[i], mod);
  }
  return res;
}
Num pow(Num x, int k) {
  Num res = x, tmp = x;
  k--;
  while (k) {
    if (k & 1) res = merge(res, tmp);
    tmp = merge(tmp, tmp);
    k >>= 1;
  }
  return res;
}
void solve() {
  cin >> n >> k;
  vector<int> a, b;
  a.push_back(1), a.push_back(1), b.push_back(1);
  Num res = pow(Num(a, b), n + 1);
  while (int((res.b).size()) < k + 1) res.b.push_back(0);
  for (int i = (1); i < (k + 1); i++)
    cout << mult(res.b[i], fpow(inv(2, mod), n, mod), mod) << " \n"[i == k];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("in.txt", "r")) {
    freopen("in.txt", "r", stdin);
  }
  solve();
  cerr << "\nTime elapsed: " << 1000.0 * clock() / CLOCKS_PER_SEC << " ms\n";
  return 0;
}