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 = 998244353;
const int G = 3, iG = 332748118, inv2 = 499122177;
const int maxn = 262144;
inline int read() {
  char c = getchar();
  int out = 0;
  while (!(c >= '0' && c <= '9')) c = getchar();
  while (c >= '0' && c <= '9') out = out * 10 + c - '0', c = getchar();
  return out;
}
inline int add(int a, int b) {
  if ((a += b) >= mod) a -= mod;
  return a;
}
inline void add_(int &a, int b) {
  if ((a += b) >= mod) a -= mod;
}
inline int dec(int a, int b) {
  if ((a -= b) < 0) a += mod;
  return a;
}
inline void dec_(int &a, int b) {
  if ((a -= b) < 0) a += mod;
}
inline int mult(int a, int b) {
  long long t = 1ll * a * b;
  if (t >= mod) t %= mod;
  return t;
}
inline int power(int a, int b) {
  int out = 1;
  while (b) {
    if (b & 1) out = mult(out, a);
    a = mult(a, a);
    b >>= 1;
  }
  return out;
}
namespace polynomial {
int up[maxn + 5];
void init_up() {
  up[0] = up[1] = 0;
  for (int i = 1; (1 << i) <= maxn; i++) {
    for (int j = ((1 << (i - 1)) + 1); j <= (1 << i); j++) up[j] = i;
  }
}
int R[maxn + 5], pw[maxn + 5], ipw[maxn + 5];
void init_pw() {
  init_up();
  for (int i = 1; i <= maxn; i++)
    pw[i] = power(G, (mod - 1) / (i << 1)),
    ipw[i] = power(iG, (mod - 1) / (i << 1));
}
void init_R(int len) {
  for (int i = 0; i < (1 << len); i++)
    R[i] = (R[i >> 1] >> 1) | ((i & 1) << (len - 1));
}
inline void dft(vector<int> &A, int len, int flag, bool reinit = 1) {
  int L = up[len];
  if (reinit) init_R(L);
  for (int i = 0; i < (1 << L); i++)
    if (i < R[i]) swap(A[i], A[R[i]]);
  for (int i = 1; i < (1 << L); i <<= 1) {
    int bas = (flag == 1) ? pw[i] : ipw[i];
    for (int j = 0; j < (1 << L); j += (i << 1)) {
      int t = 1;
      for (int k = 0; k < i; k++, t = mult(t, bas)) {
        int nx = A[j + k], ny = mult(A[i + j + k], t);
        A[j + k] = add(nx, ny);
        A[i + j + k] = dec(nx, ny);
      }
    }
  }
  if (flag == -1) {
    int iL = power(1 << L, mod - 2);
    for (int i = 0; i < (1 << L); i++) A[i] = mult(A[i], iL);
  }
}
inline vector<int> Mult(vector<int> A, vector<int> B) {
  int totlen = 1 << up[A.size() + B.size() - 1];
  A.resize(totlen);
  B.resize(totlen);
  dft(A, totlen, 1);
  dft(B, totlen, 1, 0);
  for (int i = 0; i < totlen; i++) A[i] = mult(A[i], B[i]);
  dft(A, totlen, -1, 0);
  A.resize(A.size() + B.size() - 1);
  return A;
}
inline void Mult_ref(vector<int> &A, vector<int> B) {
  int totlen = 1 << up[A.size() + B.size() - 1];
  A.resize(totlen);
  B.resize(totlen);
  dft(A, totlen, 1);
  dft(B, totlen, 1, 0);
  for (int i = 0; i < totlen; i++) A[i] = mult(A[i], B[i]);
  dft(A, totlen, -1, 0);
  A.resize(A.size() + B.size() - 1);
}
inline vector<int> Add(vector<int> A, vector<int> B) {
  vector<int> ret(max(A.size(), B.size()));
  for (int i = 0; i < ret.size(); i++) {
    if (i < A.size()) ret[i] = A[i];
    if (i < B.size()) ret[i] = add(ret[i], B[i]);
  }
  return ret;
}
inline vector<int> inverse(vector<int> A, int len) {
  vector<int> B(1, power(A[0], mod - 2));
  for (int L = 2; L <= len; L <<= 1) {
    vector<int> tA = A;
    tA.resize(L);
    tA.resize(L << 1);
    B.resize(L << 1);
    dft(tA, L << 1, 1);
    dft(B, L << 1, 1, 0);
    for (int i = 0; i < (L << 1); i++)
      B[i] = mult(B[i], dec(2, mult(tA[i], B[i])));
    dft(B, L << 1, -1, 0);
    B.resize(L);
  }
  return B;
}
inline vector<int> inverse(vector<int> A) {
  int len = A.size();
  A.resize(1 << up[A.size()]);
  vector<int> ret = inverse(A, A.size());
  ret.resize(len);
  return ret;
}
inline vector<int> derivative(vector<int> A) {
  for (int i = 0; i + 1 < A.size(); i++) A[i] = mult(i + 1, A[i + 1]);
  A.pop_back();
  return A;
}
inline vector<int> integral(vector<int> A) {
  vector<int> ret(A.size() + 1, 0);
  for (int i = 0; i < A.size(); i++)
    ret[i + 1] = mult(A[i], power(i + 1, mod - 2));
  return ret;
}
inline vector<int> ln(vector<int> A) {
  int len = A.size();
  A = integral(Mult(derivative(A), inverse(A)));
  A.resize(len);
  return A;
}
inline vector<int> exp(vector<int> A, int len) {
  vector<int> B(1, 1);
  for (int L = 2; L <= len; L <<= 1) {
    vector<int> tA = A;
    tA.resize(L);
    B.resize(L);
    vector<int> tB = ln(B);
    for (int i = 0; i < L; i++) tA[i] = dec(tA[i], tB[i]);
    tA[0] = add(tA[0], 1);
    tA.resize(L << 1);
    B.resize(L << 1);
    dft(tA, L << 1, 1);
    dft(B, L << 1, 1, 0);
    for (int i = 0; i < (L << 1); i++) B[i] = mult(B[i], tA[i]);
    dft(B, L << 1, -1, 0);
    B.resize(L);
  }
  return B;
}
inline vector<int> exp(vector<int> A) {
  int len = A.size();
  A.resize(1 << up[A.size()]);
  A = exp(A, 1 << up[A.size()]);
  A.resize(len);
  return A;
}
inline vector<int> sqrt(vector<int> A, int len) {
  vector<int> B(1, 1);
  for (int L = 2; L <= len; L <<= 1) {
    vector<int> tA = A;
    tA.resize(L);
    B.resize(L);
    vector<int> tB = Mult(tA, inverse(B));
    tB.resize(L);
    for (int i = 0; i < L; i++) B[i] = mult(add(tB[i], B[i]), inv2);
  }
  return B;
}
inline vector<int> sqrt(vector<int> A) {
  int len = A.size();
  A.resize(1 << up[A.size()]);
  A = sqrt(A, 1 << up[A.size()]);
  A.resize(len);
  return A;
}
inline vector<int> Power(vector<int> A, int K) {
  int pos = -1;
  for (int i = 0; i < A.size(); i++) {
    if (A[i]) {
      pos = i;
      break;
    }
  }
  if (pos == -1) return A;
  if (pos * K >= A.size()) return vector<int>(A.size());
  int iv = power(A[pos], mod - 2);
  vector<int> B(A.size() - pos);
  for (int i = pos; i < A.size(); i++) B[i - pos] = mult(A[i], iv);
  B = ln(B);
  for (auto &x : B) x = mult(x, K);
  B = exp(B);
  B.resize(A.size());
  int pw = power(A[pos], K);
  for (int i = A.size() - 1; i >= K * pos; i--) B[i] = mult(B[i - K * pos], pw);
  for (int i = 0; i < K * pos; i++) B[i] = 0;
  return B;
}
}  // namespace polynomial
using namespace polynomial;
int n, K;
int main() {
  init_pw();
  scanf("%d%d", &n, &K);
  vector<int> delta({1, 6, 1});
  delta.resize(K + 1);
  delta = sqrt(delta);
  vector<int> d1 = delta;
  d1[0] = add(d1[0], 1);
  d1[1] = add(d1[1], 1);
  for (int &x : d1) x = mult(x, inv2);
  vector<int> &st1 = d1;
  st1 = Power(d1, n + 1);
  st1 = Mult(st1, inverse(delta));
  for (int i = 1; i <= K; i++) printf("%d ", i > n ? 0 : st1[i]);
  return 0;
}