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;
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int mult(int a, int b) { return (long long)a * b % mod; }
vector<int> sum(const vector<int> &l, const vector<int> &r) {
  vector<int> ans(max((int)(l).size(), (int)(r).size()));
  for (int i = 0; i < int((int)(l).size()); ++i) add(ans[i], l[i]);
  for (int i = 0; i < int((int)(r).size()); ++i) add(ans[i], r[i]);
  return ans;
}
vector<int> diff(const vector<int> &l, const vector<int> &r) {
  vector<int> ans(max((int)(l).size(), (int)(r).size()));
  for (int i = 0; i < int((int)(l).size()); ++i) add(ans[i], l[i]);
  for (int i = 0; i < int((int)(r).size()); ++i) sub(ans[i], r[i]);
  return ans;
}
const double pi = acosl(-1.0);
namespace FFT {
struct com {
  double x, y;
  com(double _x = 0, double _y = 0) : x(_x), y(_y) {}
  inline com operator+(const com &c) const { return com(x + c.x, y + c.y); }
  inline com operator-(const com &c) const { return com(x - c.x, y - c.y); }
  inline com operator*(const com &c) const {
    return com(x * c.x - y * c.y, x * c.y + y * c.x);
  }
  inline com conj() const { return com(x, -y); }
};
const static int maxk = 17, maxn = (1 << maxk) + 1;
com ws[maxn];
int dp[maxn];
com rs[maxn];
int n, k;
int lastk = -1;
void fft(com *a, bool torev = 0) {
  if (lastk != k) {
    lastk = k;
    dp[0] = 0;
    for (int i = 1, g = -1; i < n; ++i) {
      if (!(i & (i - 1))) {
        ++g;
      }
      dp[i] = dp[i ^ (1 << g)] ^ (1 << (k - 1 - g));
    }
    ws[1] = com(1, 0);
    for (int two = 0; two < k - 1; ++two) {
      double alf = pi / n * (1 << (k - 1 - two));
      com cur = com(cos(alf), sin(alf));
      int p2 = (1 << two), p3 = p2 * 2;
      for (int j = p2; j < p3; ++j) {
        ws[j * 2 + 1] = (ws[j * 2] = ws[j]) * cur;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    if (i < dp[i]) {
      swap(a[i], a[dp[i]]);
    }
  }
  if (torev) {
    for (int i = 0; i < n; ++i) {
      a[i].y = -a[i].y;
    }
  }
  for (int len = 1; len < n; len <<= 1) {
    for (int i = 0; i < n; i += len) {
      int wit = len;
      for (int it = 0, j = i + len; it < len; ++it, ++i, ++j) {
        com tmp = a[j] * ws[wit++];
        a[j] = a[i] - tmp;
        a[i] = a[i] + tmp;
      }
    }
  }
}
com a[maxn];
int mult(int na, const int *_a, int nb, const int *_b, long long *ans) {
  if (!na || !nb) {
    return 0;
  }
  for (k = 0, n = 1; n < na + nb - 1; n <<= 1, ++k)
    ;
  assert(n < maxn);
  for (int i = 0; i < n; ++i) {
    a[i] = com(i < na ? _a[i] : 0, i < nb ? _b[i] : 0);
  }
  fft(a);
  a[n] = a[0];
  for (int i = 0; i <= n - i; ++i) {
    a[i] = (a[i] * a[i] - (a[n - i] * a[n - i]).conj()) *
           com(0, (double)-1 / n / 4);
    a[n - i] = a[i].conj();
  }
  fft(a, 1);
  int res = 0;
  for (int i = 0; i < n; ++i) {
    long long val = (long long)round(a[i].x);
    assert(abs(val - a[i].x) < 1e-1);
    if (val) {
      assert(i < na + nb - 1);
      while (res < i) {
        ans[res++] = 0;
      }
      ans[res++] = val;
    }
  }
  return res;
}
};  // namespace FFT
vector<int> fftmult(vector<int> &l, vector<int> &r) {
  vector<long long> res((int)(l).size() + (int)(r).size() - 1);
  FFT::mult((int)(l).size(), &l[0], (int)(r).size(), &r[0], &res[0]);
  vector<int> rres((int)(res).size());
  for (int i = 0; i < int((int)(res).size()); ++i) rres[i] = res[i] % mod;
  return rres;
}
vector<int> mult_brut(const vector<int> &l, const vector<int> &r, int k) {
  vector<int> ans(k);
  for (int i = 0; i < int((int)(l).size()); ++i)
    if (l[i]) {
      int bnd = min((int)(r).size(), k - i);
      for (int j = 0; j < int(bnd); ++j) add(ans[i + j], mult(l[i], r[j]));
    }
  return ans;
}
vector<int> mult(const vector<int> &l, const vector<int> &r) {
  const int base = (int)sqrt(mod);
  vector<int> res((int)(l).size() + (int)(r).size() - 1);
  for (int tl = 0; tl < int(2); ++tl)
    for (int tr = 0; tr < int(2); ++tr) {
      vector<int> cl((int)(l).size()), cr((int)(r).size());
      int resmult = mult((tl ? base : 1), (tr ? base : 1));
      for (int i = 0; i < int((int)(l).size()); ++i)
        cl[i] = (tl ? l[i] / base : l[i] % base);
      for (int i = 0; i < int((int)(r).size()); ++i)
        cr[i] = (tr ? r[i] / base : r[i] % base);
      vector<int> curres = fftmult(cl, cr);
      curres.resize((int)(res).size());
      for (int i = 0; i < int((int)(res).size()); ++i)
        add(res[i], mult(curres[i], resmult));
    }
  return res;
}
vector<int> mult(const vector<int> &l, const vector<int> &r, int k) {
  vector<int> ans = mult(l, r);
  ans.resize(k);
  return ans;
}
vector<vector<int> > multmod(const vector<vector<int> > &l,
                             const vector<vector<int> > &r,
                             const vector<vector<int> > &bymod, int k) {
  assert((int)(l).size() == 2 && (int)(r).size() == 2 &&
         (int)(bymod).size() == 3);
  vector<vector<int> > ans(3);
  vector<int> ans00 = mult(l[0], r[0], k);
  vector<int> ans11 = mult(l[1], r[1], k);
  vector<int> ansss = mult(sum(l[0], l[1]), sum(r[0], r[1]), k);
  ans[0] = ans00;
  ans[2] = ans11;
  ans[1] = diff(ansss, sum(ans00, ans11));
  for (int i = (int)(ans).size() - 1; i >= (int)(bymod).size() - 1; i--) {
    const vector<int> &coef = ans[i];
    for (int j = 0; j < int((int)(bymod).size()); ++j) {
      int to = i - ((int)(bymod).size() - 1) + j;
      ans[to] = diff(ans[to], mult_brut(coef, bymod[j], k));
    }
    ans.pop_back();
  }
  return ans;
}
vector<vector<int> > powmod(vector<vector<int> > what, int pw,
                            const vector<vector<int> > &bymod, int k) {
  assert((int)(what).size() == 2 && (int)(bymod).size() == 3);
  vector<vector<int> > res = {{1}, {0}};
  while (pw) {
    if (pw & 1) res = multmod(res, what, bymod, k);
    what = multmod(what, what, bymod, k);
    pw >>= 1;
  }
  assert((int)(res).size() == 2);
  return res;
}
void solve(int n) {
  int k;
  cin >> k;
  k++;
  vector<vector<int> > chpoly(3);
  const int m1 = mod - 1;
  chpoly[2] = {1};
  chpoly[1] = {m1, m1};
  chpoly[0] = {0, m1};
  vector<vector<int> > poly = {{0}, {1}};
  poly = powmod(poly, n, chpoly, k);
  assert((int)(poly).size() == 2);
  vector<int> ans = sum(mult({1, 1}, poly[1], k), poly[0]);
  assert((int)(ans).size() == k);
  for (int i = 0; i < int(k); ++i)
    if (i != 0) cout << ans[i] << " \n"[i + 1 == k];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  while (cin >> n) solve(n);
}