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 maxn = 1e5 + 1;
int tail[maxn];
const long double PI = 3.141592653589793;
const int MOD = 998244353;
int root_pw = (1 << 23);
inline int pw(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = (long long)res * a % MOD;
    b >>= 1;
    a = (long long)a * a % MOD;
  }
  return res;
}
int root1 = pw(3, 119);
int inv_root = pw(root1, MOD - 2);
int cur = -1;
int iter = 0;
int* w;
int* iw;
int* cur_w;
int def[4][maxn];
void prep(int n) {
  def[0][0] = 1;
  def[1][0] = 1;
  def[1][1] = 1;
  def[2][0] = 1;
  def[2][1] = 3;
  def[2][2] = 1;
  def[3][0] = 1;
  def[3][1] = 5;
  def[3][2] = 5;
  def[3][3] = 1;
  w = new int[4 * maxn];
  iw = new int[4 * maxn];
  for (int l = 2; l <= n; l <<= 1) {
    int half = l >> 1;
    int rt = pw(3, (MOD - 1) / l);
    int inv_rt = pw(rt, MOD - 2);
    int cur_rt = 1;
    int cur_irt = 1;
    for (int i = 0; i < half; ++i, ++iter) {
      w[iter] = cur_rt;
      iw[iter] = cur_irt;
      cur_rt = (long long)cur_rt * rt % MOD;
      cur_irt = (long long)cur_irt * inv_rt % MOD;
    }
  }
}
int* fft(int n, int* x, bool invert) {
  int k = 0;
  while ((1 << k) < n) ++k;
  n = (1 << k);
  cur_w = invert ? iw : w;
  int* a = new int[n];
  memcpy(a, x, sizeof(int) * n);
  if (cur != n) {
    for (int i = 0; i < n; ++i) {
      tail[i] = (i & 1) << (k - 1) | (tail[i >> 1] >> 1);
      cur = n;
    }
  }
  for (int i = 0; i < n; ++i)
    if (i < tail[i]) swap(a[i], a[tail[i]]);
  for (int len = 2; len <= n; len <<= 1) {
    int half = len >> 1;
    for (int bl = 0; bl < n; bl += len) {
      int pos = half - 1;
      for (int i = bl; i < bl + half; ++i, ++pos) {
        int j = i + half;
        int fi = a[i];
        int se = (long long)cur_w[pos] * a[j] % MOD;
        a[i] = fi + se;
        if (a[i] >= MOD) a[i] -= MOD;
        a[j] = fi - se;
        if (a[j] < 0) a[j] += MOD;
      }
    }
  }
  if (invert) {
    int inv = pw(n, MOD - 2);
    for (int i = 0; i < n; ++i) a[i] = (long long)a[i] * inv % MOD;
  }
  return a;
}
int sz;
int* mult(int* a, int* b, int n) {
  int k = 0;
  while ((1 << k) < n) ++k;
  n = (1 << k);
  n <<= 1;
  int* a1 = fft(n, a, 0);
  int* b1 = fft(n, b, 0);
  for (int i = 0; i < n; ++i) a1[i] = (long long)a1[i] * b1[i] % MOD;
  a1 = fft(n, a1, 1);
  delete[] b1;
  return a1;
}
int layer[3][maxn];
void bin(int n) {
  if (n <= 2) {
    memcpy(layer[0], def[n], sz * sizeof(int));
    memcpy(layer[1], def[n - 1], sz * sizeof(int));
    memcpy(layer[2], def[n - 2], sz * sizeof(int));
    return;
  }
  if (n & 1) {
    bin(n - 1);
    int* fi = mult(layer[0], def[1], sz);
    int* temp = mult(layer[1], def[0], sz);
    for (int i = 0; i < sz - 1; ++i) {
      fi[i + 1] += temp[i];
      if (fi[i + 1] >= MOD) fi[i + 1] -= MOD;
    }
    memcpy(layer[2], layer[1], sz * sizeof(int));
    memcpy(layer[1], layer[0], sz * sizeof(int));
    memcpy(layer[0], fi, sz * sizeof(int));
    delete[] fi;
    delete[] temp;
  } else {
    bin(n / 2);
    int* fi = mult(layer[0], layer[0], sz);
    int* temp = mult(layer[1], layer[1], sz);
    int* th = temp;
    for (int i = 0; i < sz - 1; ++i) {
      fi[i + 1] += temp[i];
      if (fi[i + 1] >= MOD) fi[i + 1] -= MOD;
    }
    int* se = mult(layer[0], layer[1], sz);
    temp = mult(layer[1], layer[2], sz);
    for (int i = 0; i < sz - 1; ++i) {
      se[i + 1] += temp[i];
      if (se[i + 1] >= MOD) se[i + 1] -= MOD;
    }
    temp = mult(layer[2], layer[2], sz);
    for (int i = 0; i < sz - 1; ++i) {
      th[i + 1] += temp[i];
      if (th[i + 1] >= MOD) th[i + 1] -= MOD;
    }
    memcpy(layer[0], fi, sz * sizeof(int));
    memcpy(layer[1], se, sz * sizeof(int));
    memcpy(layer[2], th, sz * sizeof(int));
    delete[] temp;
    delete[] fi;
    delete[] se;
    delete[] th;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int k;
  scanf("%d", &k);
  sz = max((k + 1), 10);
  int lg = 0;
  while ((1 << lg) < sz) ++lg;
  sz = (1 << lg);
  prep(2 * sz);
  if (n <= 2) {
    for (int i = 1; i <= k; ++i) printf("%d ", def[n][i]);
    return 0;
  }
  bin(n);
  for (int i = 1; i <= k; ++i) printf("%d ", layer[0][i]);
}