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;
const int oo = 0x3f3f3f3f;
const long long ooo = 9223372036854775807ll;
const int _cnt = 1000 * 1000 + 7;
const int _p = 998244353;
const int N = 200005;
const double PI = acos(-1.0);
const double eps = 1e-9;
int o(int x) { return x % _p; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
void file_put() {
  freopen("filename.in", "r", stdin);
  freopen("filename.out", "w", stdout);
}
const int _N = 200005;
long long inv[_N << 2];
inline long long add(long long x, long long y) {
  x += y;
  return x % _p;
}
inline long long mul(long long x, long long y) { return (long long)x * y % _p; }
inline long long Pow(long long x, long long k) {
  long long ans = 1;
  for (; k; k >>= 1, x = x * x % _p)
    if (k & 1) (ans *= x) %= _p;
  return ans;
}
inline void init_inv(int n) {
  inv[1] = 1;
  for (int i = (2); i <= (n); ++i) inv[i] = mul(_p - _p / i, inv[_p % i]);
}
template <class V>
struct FT {
  int n, nn;
  V w[2][_N << 2], rev[_N << 2], tmp;
  inline int init_len(int _n) {
    for (n = 1; n <= _n; n <<= 1)
      ;
    return n;
  }
  inline int Init(int _n) {
    init_len(_n);
    if (n == nn) return n;
    nn = n;
    V w0 = Pow(3, (_p - 1) / n);
    w[0][0] = w[1][0] = 1;
    for (int i = (1); i <= (n - 1); ++i)
      w[0][i] = w[1][n - i] = mul(w[0][i - 1], w0);
    for (int i = (0); i <= (n - 1); ++i)
      rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (n >> 1));
    return n;
  }
  void FFT(V A[], int op) {
    for (int i = (0); i <= (n - 1); ++i)
      if (i < rev[i]) swap(A[i], A[rev[i]]);
    for (int i = 1; i < n; i <<= 1)
      for (int j = 0, t = n / (i << 1); j < n; j += i << 1)
        for (int k = j, l = 0; k < j + i; k++, l += t) {
          V x = A[k], y = mul(w[op][l], A[k + i]);
          A[k] = add(x, y), A[k + i] = add(x - y, _p);
        }
    if (op) {
      tmp = inv[n];
      for (int i = (0); i <= (n - 1); ++i) A[i] = mul(A[i], tmp);
    }
  }
};
template <class V>
struct Calculator {
  FT<V> T;
  V X[_N << 2], Y[_N << 2], A[_N << 2], B[_N << 2], C[_N << 2];
  inline void Fill(V a[], V b[], int n, int len) {
    if (a != b) memcpy(a, b, sizeof(V) * n);
    fill(a + n, a + len, 0);
  }
  inline void Add(V a[], int n, V b[], int m, V c[], int t = 1) {
    n = ((n) > (m) ? (n) : (m));
    for (int i = (0); i <= (n - 1); ++i) c[i] = add(a[i], t * b[i]);
  }
  inline void Dot_Mul(V a[], V b[], int len, V c[]) {
    for (int i = (0); i <= (len - 1); ++i) c[i] = mul(a[i], b[i]);
  }
  inline void Dot_Mul(V a[], int len, V v, V c[]) {
    for (int i = (0); i <= (len - 1); ++i) c[i] = mul(a[i], v);
  }
  inline void Mul(V a[], int n, V b[], int m, V c[]) {
    int len = T.Init(n + m - 1);
    Fill(X, a, n, len), Fill(Y, b, m, len);
    T.FFT(X, 0), T.FFT(Y, 0), Dot_Mul(X, Y, len, c), T.FFT(c, 1);
  }
  inline void Int(V a[], int n, V b[]) {
    for (int i = (n - 1); i >= (0); --i) b[i + 1] = mul(a[i], inv[i + 1]);
    b[0] = 0;
  }
  inline void Der(V a[], int n, V b[]) {
    for (int i = (1); i <= (n - 1); ++i) b[i - 1] = mul(a[i], i);
    b[n - 1] = 0;
  }
  inline void Inv(V a[], int n, V b[]) {
    if (n == 1) {
      b[0] = Pow(a[0], _p - 2), b[1] = 0;
      return;
    }
    Inv(a, (n + 1) >> 1, b);
    int len = T.Init(2 * n - 1);
    Fill(X, a, n, len), Fill(b, b, n, len), T.FFT(X, 0), T.FFT(b, 0);
    for (int i = (0); i <= (len - 1); ++i)
      b[i] = mul(b[i], 2 - mul(b[i], X[i]));
    T.FFT(b, 1), Fill(b, b, n, len);
  }
  inline void Log(V a[], int n, V b[]) {
    static V A[_N << 2], B[_N << 2];
    Der(a, n, A), Inv(a, n, B), Mul(A, n, B, n, b);
    Int(b, n, b), Fill(b, b, n, T.n);
  }
  inline void Exp(V a[], int n, V b[]) {
    if (n == 1) {
      b[0] = exp(a[0]), b[1] = 0;
      return;
    }
    Exp(a, (n + 1) >> 1, A), Log(A, n, B), Add(a, n, B, n, B, -1);
    (B[0] += 1) %= _p, Mul(A, n, B, n, b), Fill(b, b, n, T.n);
  }
  inline void Sqrt(V a[], int n, V b[]) {
    if (n == 1) {
      b[0] = sqrt(a[0]), b[1] = 0;
      return;
    }
    Sqrt(a, (n + 1) >> 1, b), Inv(b, n, B), Mul(a, n, B, n, B);
    Add(b, n, B, n, b), Dot_Mul(b, n, inv[2], b), Fill(b, b, n, T.n);
  }
  inline void Power(V a[], int n, long long k, V b[]) {
    Log(a, n, C), Dot_Mul(C, n, k, C), Exp(C, n, b), Fill(b, b, n, T.n);
  }
};
int n, k, kk;
long long D[N], A[N], B[N], E[N];
Calculator<long long> T;
int main() {
  scanf("%d%d", &n, &kk), k = ((kk) < (n) ? (kk) : (n)) + 1, init_inv(N - 1);
  E[0] = 1, E[1] = 6, E[2] = 1;
  T.Sqrt(E, k, D);
  T.Fill(A, D, k, k);
  A[0]++, A[1]++;
  long long x = A[0];
  T.Dot_Mul(A, k, Pow(x, _p - 2), A);
  T.Power(A, k, n + 1, B), T.Dot_Mul(B, k, Pow(x, n + 1), B), T.Inv(D, k, E);
  T.Mul(B, k, E, k, A), T.Dot_Mul(A, k, Pow(_p + 1 >> 1, n + 1), A),
      T.Fill(A, A, k, T.T.n);
  for (int i = (1); i <= (kk); ++i) printf("%lld ", A[i]);
  printf("\n");
  return 0;
}
