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>
#pragma comment(linker, "/STACK:64000000")
using namespace std;
const int MAXK = -1;
const int MAXN = -1;
const int MOD = 998244353;
namespace FFT {
const int maxBase = 21;
const int maxN = 1 << maxBase;
struct num {
  double x, y;
  num() {}
  num(double xx, double yy) : x(xx), y(yy) {}
  num(double alp) : x(cos(alp)), y(sin(alp)) {}
};
inline num operator+(num a, num b) { return num(a.x + b.x, a.y + b.y); }
inline num operator-(num a, num b) { return num(a.x - b.x, a.y - b.y); }
inline num operator*(num a, num b) {
  return num(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
}
inline num conj(num a) { return num(a.x, -a.y); }
const double PI = acos(-1);
num root[maxN];
int rev[maxN];
bool rootsPrepared = false;
void prepRoots() {
  if (rootsPrepared) return;
  rootsPrepared = true;
  root[1] = num(1, 0);
  for (int k = 1; k < maxBase; ++k) {
    num x(2 * PI / (1LL << (k + 1)));
    for (int i = (1LL << (k - 1)); i < (1LL << (k)); ++i) {
      root[2 * i] = root[i];
      root[2 * i + 1] = root[i] * x;
    }
  }
}
int base, N;
int lastRevN = -1;
void prepRev() {
  if (lastRevN == N) return;
  lastRevN = N;
  for (int i = 0; i < (N); i++)
    rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (base - 1));
}
void fft(num *a, num *f) {
  for (int i = 0; i < (N); i++) f[i] = a[rev[i]];
  for (int k = 1; k < N; k <<= 1)
    for (int i = 0; i < N; i += 2 * k)
      for (int j = 0; j < (k); j++) {
        num z = f[i + j + k] * root[j + k];
        f[i + j + k] = f[i + j] - z;
        f[i + j] = f[i + j] + z;
      }
}
num a[maxN], b[maxN], f[maxN], g[maxN];
long long A[maxN], B[maxN], C[maxN];
void _multMod(int mod) {
  for (int i = 0; i < (N); i++) {
    int x = A[i] % mod;
    a[i] = num(x & ((1LL << (15)) - 1), x >> 15);
  }
  for (int i = 0; i < (N); i++) {
    int x = B[i] % mod;
    b[i] = num(x & ((1LL << (15)) - 1), x >> 15);
  }
  fft(a, f);
  fft(b, g);
  for (int i = 0; i < (N); i++) {
    int j = (N - i) & (N - 1);
    num a1 = (f[i] + conj(f[j])) * num(0.5, 0);
    num a2 = (f[i] - conj(f[j])) * num(0, -0.5);
    num b1 = (g[i] + conj(g[j])) * num(0.5 / N, 0);
    num b2 = (g[i] - conj(g[j])) * num(0, -0.5 / N);
    a[j] = a1 * b1 + a2 * b2 * num(0, 1);
    b[j] = a1 * b2 + a2 * b1;
  }
  fft(a, f);
  fft(b, g);
  for (int i = 0; i < (N); i++) {
    long long aa = f[i].x + 0.5;
    long long bb = g[i].x + 0.5;
    long long cc = f[i].y + 0.5;
    C[i] = (aa + bb % mod * (1LL << (15)) + cc % mod * (1LL << (30))) % mod;
  }
}
void prepAB(int n1, int n2) {
  base = 1;
  N = 2;
  while (N < n1 + n2) base++, N <<= 1;
  for (int i = n1; i < N; ++i) A[i] = 0;
  for (int i = n2; i < N; ++i) B[i] = 0;
  prepRoots();
  prepRev();
}
void mult(int n1, int n2) {
  prepAB(n1, n2);
  for (int i = 0; i < (N); i++) a[i] = num(A[i], B[i]);
  fft(a, f);
  for (int i = 0; i < (N); i++) {
    int j = (N - i) & (N - 1);
    a[i] = (f[j] * f[j] - conj(f[i] * f[i])) * num(0, -0.25 / N);
  }
  fft(a, f);
  for (int i = 0; i < (N); i++) C[i] = (long long)round(f[i].x);
}
void multMod(int n1, int n2, int mod) {
  prepAB(n1, n2);
  _multMod(mod);
}
int D[maxN];
void multLL(int n1, int n2) {
  prepAB(n1, n2);
  int mod1 = 1.5e9;
  int mod2 = mod1 + 1;
  _multMod(mod1);
  for (int i = 0; i < (N); i++) D[i] = C[i];
  _multMod(mod2);
  for (int i = 0; i < (N); i++) {
    C[i] =
        D[i] + (C[i] - D[i] + (long long)mod2) * (long long)mod1 % mod2 * mod1;
  }
}
}  // namespace FFT
map<int, vector<int> > mp;
vector<int> solve(int n, int k) {
  if (mp.count(n)) return mp[n];
  if (n == 0) {
    vector<int> res(k + 1, 0);
    res[0] = 1;
    return mp[n] = res;
  }
  if (n == 1) {
    vector<int> res(k + 1, 0);
    res[0] = 1;
    res[1] = 1;
    return mp[n] = res;
  }
  vector<int> o1 = solve(n / 2, k);
  for (int i = 0; i <= k; i++) {
    FFT::A[i] = FFT::B[i] = o1[i];
  }
  FFT::multMod(k + 1, k + 1, MOD);
  vector<int> res1(k + 1);
  for (int i = 0; i <= k; i++) res1[i] = FFT::C[i];
  vector<int> o2 = solve(n / 2 - (n % 2 == 0), k);
  for (int i = 0; i <= k; i++) {
    FFT::A[i] = FFT::B[i] = o2[i];
  }
  FFT::multMod(k + 1, k + 1, MOD);
  vector<int> res2(k + 1);
  for (int i = 1; i <= k; i++) res2[i] = FFT::C[i - 1];
  vector<int> res(k + 1);
  for (int i = 0; i <= k; i++) res[i] = (res1[i] + res2[i]) % MOD;
  if (n % 2 == 1) {
    vector<int> o1 = solve(n / 2, k);
    vector<int> o2 = solve(n / 2 - 1, k);
    for (int i = 0; i < k; i++) {
      FFT::A[i] = o1[i];
      FFT::B[i] = o2[i];
    }
    FFT::multMod(k + 1, k + 1, MOD);
    vector<int> res3(k + 1);
    for (int i = 1; i <= k; i++) res3[i] = FFT::C[i - 1];
    for (int i = 0; i <= k; i++) res[i] = (res[i] + 2LL * res3[i]) % MOD;
  }
  return mp[n] = res;
}
int main() {
  int n, k;
  while (cin >> n >> k) {
    mp.clear();
    vector<int> ans = solve(n, k);
    for (int i = 1; i <= k; i++) printf("%d%c", ans[i], " \n"[i == k]);
  }
  return 0;
}