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;
int Read() {
  int x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x;
}
void Write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x == 0) putchar('0');
  int stk[55], tp = 0;
  while (x) stk[++tp] = x % 10, x /= 10;
  for (int i = tp; i; i--) putchar(stk[i] + '0');
}
int Add(int a, int b) {
  return (a + b >= 998244353) ? a + b - 998244353 : a + b;
}
int Dec(int a, int b) { return (a - b < 0) ? a - b + 998244353 : a - b; }
int mul(int a, int b) { return 1ll * a * b % 998244353; }
unsigned int qpow(unsigned int a, int b) {
  unsigned int res = 1;
  for (; b; b >>= 1, a = mul(a, a))
    if (b & 1) res = mul(res, a);
  return res;
}
namespace NTT {
int sz;
unsigned int w[2500005], w_mf[2500005];
int mf(int x) { return (1ll * x << 32) / 998244353; }
void init(int n) {
  for (sz = 2; sz < n; sz <<= 1)
    ;
  unsigned int pr = qpow(3, (998244353 - 1) / sz);
  w[sz / 2] = 1;
  w_mf[sz / 2] = mf(1);
  for (int i = 1; i <= sz / 2; i++)
    w[sz / 2 + i] = mul(w[sz / 2 + i - 1], pr),
               w_mf[sz / 2 + i] = mf(w[sz / 2 + i]);
  for (int i = sz / 2 - 1; i; i--) w[i] = w[i << 1], w_mf[i] = w_mf[i << 1];
}
void ntt(vector<unsigned int>& A, int L) {
  for (int d = L >> 1; d; d >>= 1)
    for (int i = 0; i < L; i += (d << 1))
      for (int j = 0; j < d; j++) {
        unsigned int x = A[i + j] + A[i + d + j];
        if (x >= 998244353 * 2) x -= 998244353 * 2;
        long long t = A[i + j] + 998244353 * 2 - A[i + d + j],
                  q = t * w_mf[d + j] >> 32, y = t * w[d + j] - q * 998244353;
        A[i + j] = x;
        A[i + d + j] = y;
      }
  for (int i = 0; i < L; i++)
    if (A[i] >= 998244353) A[i] -= 998244353;
}
void intt(vector<unsigned int>& A, int L) {
  for (int d = 1; d < L; d <<= 1)
    for (int i = 0; i < L; i += (d << 1))
      for (int j = 0; j < d; j++) {
        unsigned int x = A[i + j];
        if (x >= 998244353 * 2) x -= 998244353 * 2;
        long long t = A[i + d + j], q = t * w_mf[d + j] >> 32,
                  y = t * w[d + j] - q * 998244353;
        A[i + j] = x + y, A[i + d + j] = x + 2 * 998244353 - y;
      }
  int k = __builtin_ctz(L);
  reverse(A.begin() + 1, A.end());
  for (int i = 0; i < L; i++) {
    long long m = -A[i] & (L - 1);
    A[i] = (A[i] + m * 998244353) >> k;
    if (A[i] >= 998244353) A[i] -= 998244353;
  }
}
}  // namespace NTT
struct poly {
  vector<unsigned int> a;
  poly(int d = 0, int t = 0) {
    a.resize(d + 1);
    a[d] = t;
  }
  int deg() { return a.size() - 1; }
  unsigned int& operator[](const int& b) { return a[b]; }
  poly extend(int x) {
    poly c = *this;
    c.a.resize(x + 1);
    return c;
  }
  void resize(int x) { a.resize(x); }
  int size() { return a.size(); }
  void Get(int n) {
    a.resize(n + 1);
    for (int i = 0; i <= n; i++) a[i] = Read();
  }
  void Print(int n) {
    for (int i = 0; i <= n; i++) Write(a[i]), putchar(' ');
    puts("");
  }
  friend poly operator*(poly A, poly B) {
    int n = A.deg() + B.deg() + 1, lim = 2;
    for (; lim < n; lim <<= 1)
      ;
    NTT::init(lim);
    A.resize(lim);
    B.resize(lim);
    NTT::ntt(A.a, lim);
    NTT::ntt(B.a, lim);
    for (int i = 0; i < lim; i++) A[i] = mul(A[i], B[i]);
    NTT::intt(A.a, lim);
    return A.extend(n - 1);
  }
  poly inv(poly A) {
    int n = A.a.size(), lim = 2;
    if (n == 1) return A[0] = qpow(A[0], 998244353 - 2), A;
    poly B = A;
    B.resize((n + 1) >> 1);
    B = inv(B);
    for (; lim < (n << 1); lim <<= 1)
      ;
    NTT::init(lim);
    A.resize(lim), B.resize(lim), NTT::ntt(A.a, lim);
    NTT::ntt(B.a, lim);
    for (int i = 0; i < lim; i++) A[i] = mul(Dec(2, mul(A[i], B[i])), B[i]);
    NTT::intt(A.a, lim);
    return A.extend(n - 1);
  }
  poly getinv() { return inv(*this); }
  poly getdao() {
    poly A = *this, B;
    int n = A.a.size();
    B.resize(n);
    for (int i = 1; i < n; i++) B[i - 1] = mul(A[i], i);
    B[n - 1] = 0;
    return B;
  }
  poly getjf() {
    poly A = *this, B;
    int n = A.a.size();
    B.resize(n);
    for (int i = 1; i < n; i++) B[i] = mul(A[i - 1], qpow(i, 998244353 - 2));
    B[0] = 0;
    return B;
  }
  poly getln() {
    poly A = *this;
    return (A.getdao() * A.getinv()).extend(A.deg()).getjf();
  }
  poly exp(poly A) {
    int n = A.a.size();
    if (n == 1) return A[0] = 1, A;
    poly B = A;
    B.resize((n + 1) >> 1);
    B = exp(B);
    poly C = B.getln();
    C.resize(n);
    for (int i = 0; i < n; i++) C[i] = Dec(A[i], C[i]);
    ++C[0];
    return (B * C).extend(n - 1);
  }
  poly getexp() { return exp(*this); }
  poly getmi(int k) {
    poly A = *this;
    int n = A.a.size();
    poly B = A.getln();
    for (int i = 0; i < n; i++) B[i] = mul(B[i], k);
    return B.getexp();
  }
  poly sqrt(poly A) {
    int n = A.size(), lim = 2;
    if (n == 1) return A[0] = 1, A;
    poly B = A;
    B.resize((n + 1) >> 1);
    B = sqrt(B);
    poly C = A.extend(n - 1), D = B.getinv();
    for (; lim < (n << 1); lim <<= 1)
      ;
    NTT::init(lim);
    B.resize(lim), C.resize(lim), D.resize(lim), NTT::ntt(B.a, lim);
    NTT::ntt(C.a, lim);
    NTT::ntt(D.a, lim);
    for (int i = 0; i < lim; i++)
      B[i] = mul(Add(B[i], mul(C[i], D[i])), 499122177);
    NTT::intt(B.a, lim);
    return B.extend(n - 1);
  }
  poly getsqrt() { return sqrt(*this); }
};
signed main() {
  int n = Read(), k = Read();
  poly A, B;
  A.resize(k * 3), B.resize(k * 3);
  A[0] = 1;
  A[1] = 6;
  A[2] = 1;
  A = A.getsqrt();
  B = A.getinv();
  ++A[1];
  ++A[0];
  for (int i = 0; i <= k; i++) A[i] = 1ll * A[i] * 499122177 % 998244353;
  A = A.getmi(n + 1);
  A = (A * B).extend(k * 3);
  for (int i = 1; i <= k; i++) {
    if (i <= n)
      printf("%d ", A[i]);
    else
      printf("%d ", 0);
  }
  return 0;
}