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 = (1 << 18) + 7;
const int maxLen = 18, maxm = 1 << maxLen | 1;
const long long maxv = 1e18 + 6;
const double pi = acos(-1.0);
long long mod = ((7 * 17) << 23) + 1, nlim, sp, msk;
long long qpow(long long x, long long p) {
  long long ret = 1;
  while (p) {
    if (p & 1) (ret *= x) %= mod;
    (x *= x) %= mod;
    p >>= 1;
  }
  return ret;
}
inline int add(int a, int b) { return (a + b) % mod; }
inline int sub(int a, int b) { return (a - b + mod) % mod; }
inline int mul(int a, int b) { return (long long)a * (long long)b % mod; }
namespace FFT {
long long I2 = qpow(2, mod - 2);
struct cp {
  double r, i;
  cp() {}
  cp(double r, double i) : r(r), i(i) {}
  cp operator+(cp const &t) const { return cp(r + t.r, i + t.i); }
  cp operator-(cp const &t) const { return cp(r - t.r, i - t.i); }
  cp operator*(cp const &t) const {
    return cp(r * t.r - i * t.i, r * t.i + i * t.r);
  }
  cp conj() const { return cp(r, -i); }
} w[maxm];
void init() {
  for (int i = 0, ilim = 1 << maxLen; i < ilim; ++i) {
    int j = i, k = ilim >> 1;
    for (; !(j & 1) && !(k & 1); j >>= 1, k >>= 1)
      ;
    w[i] = cp(cos(pi / k * j), sin(pi / k * j));
  }
  nlim = std::min(maxv / (mod - 1) / (mod - 1), maxn - 1LL);
  for (sp = 1; 1 << (sp << 1) < mod; ++sp)
    ;
  msk = (1 << sp) - 1;
}
void FFT(int n, cp a[], int flag) {
  static int bitLen = 0, bitRev[maxm] = {};
  if (n != (1 << bitLen)) {
    for (bitLen = 0; 1 << bitLen < n; ++bitLen)
      ;
    for (int i = 1; i < n; ++i)
      bitRev[i] = (bitRev[i >> 1] >> 1) | ((i & 1) << (bitLen - 1));
  }
  for (int i = 0; i < n; ++i)
    if (i < bitRev[i]) std::swap(a[i], a[bitRev[i]]);
  for (int i = 1, d = 1; d < n; ++i, d <<= 1)
    for (int j = 0; j < n; j += d << 1)
      for (int k = 0; k < d; ++k) {
        cp &AL = a[j + k], &AH = a[j + k + d];
        cp TP = w[k << (maxLen - i)] * AH;
        AH = AL - TP, AL = AL + TP;
      }
  if (flag != -1) return;
  std::reverse(a + 1, a + n);
  for (int i = 0; i < n; ++i) {
    a[i].r /= n;
    a[i].i /= n;
  }
}
void polyMul(int a[], int aLen, int b[], int bLen, int c[]) {
  static cp A[maxm], B[maxm], C[maxm], D[maxm];
  int len, cLen = aLen + bLen - 1;
  for (len = 1; len < aLen + bLen - 1; len <<= 1)
    ;
  if (std::min(aLen, bLen) <= nlim) {
    for (int i = 0; i < len; ++i)
      A[i] = cp(i < aLen ? a[i] : 0, i < bLen ? b[i] : 0);
    FFT(len, A, 1);
    cp tr(0, -0.25);
    for (int i = 0, j; i < len; ++i)
      j = (len - i) & (len - 1),
      B[i] = (A[i] * A[i] - (A[j] * A[j]).conj()) * tr;
    FFT(len, B, -1);
    for (int i = 0; i < cLen; ++i) c[i] = (long long)(B[i].r + 0.5) % mod;
    return;
  }
  for (int i = 0; i < len; ++i) {
    A[i] = i < aLen ? cp(a[i] & msk, a[i] >> sp) : cp(0, 0);
    B[i] = i < bLen ? cp(b[i] & msk, b[i] >> sp) : cp(0, 0);
  }
  FFT(len, A, 1), FFT(len, B, 1);
  cp trL(0.5, 0), trH(0, -0.5), tr(0, 1);
  for (int i = 0, j; i < len; ++i) {
    j = (len - i) & (len - 1);
    cp AL = (A[i] + A[j].conj()) * trL;
    cp AH = (A[i] - A[j].conj()) * trH;
    cp BL = (B[i] + B[j].conj()) * trL;
    cp BH = (B[i] - B[j].conj()) * trH;
    C[i] = AL * (BL + BH * tr);
    D[i] = AH * (BL + BH * tr);
  }
  FFT(len, C, -1), FFT(len, D, -1);
  for (int i = 0; i < cLen; ++i) {
    int v11 = (long long)(C[i].r + 0.5) % mod,
        v12 = (long long)(C[i].i + 0.5) % mod;
    int v21 = (long long)(D[i].r + 0.5) % mod,
        v22 = (long long)(D[i].i + 0.5) % mod;
    c[i] = (((((long long)v22 << sp) + v12 + v21) << sp) + v11) % mod;
  }
}
void polyInv(int x[], int y[], int deg) {
  if (deg == 1) {
    y[0] = qpow(x[0], mod - 2);
    return;
  }
  static int c[maxm], tmp[maxm];
  polyInv(x, y, (deg + 1) >> 1);
  copy(x, x + deg, tmp);
  int p = ((deg + 1) >> 1) + deg - 1;
  polyMul(y, (deg + 1) >> 1, tmp, deg, c);
  for (int i = 0; i < p; i += 1) c[i] = (mod - c[i]) % mod;
  (c[0] += 2) %= mod;
  polyMul(y, (deg + 1) >> 1, c, deg, tmp);
  copy(tmp, tmp + deg, y);
}
void polySqrt(int A[], int B[], int deg) {
  static int T[maxm], B2[maxm];
  fill(B, B + (deg << 1), 0);
  B[0] = 1;
  for (int i = 4, j = 2; j <= deg; j = i, i <<= 1) {
    polyInv(B, T, j);
    polyMul(B, j, B, j, B2);
    for (int k = 0; k < j; k += 1) B2[k] = (B2[k] + A[k]) % mod;
    polyMul(B2, j, T, j, B);
    for (int k = 0; k < j; k += 1) B[k] = (long long)B[k] * I2 % mod;
    fill(B + j, B + i, 0);
  }
}
void polyDiv(int A[], int B[], int D[], int R[], int n, int m) {
  static int AR[maxm], BR[maxm], tmp[maxm];
  int p = n - m + 1;
  fill(AR, AR + p, 0);
  reverse_copy(B, B + m, AR);
  polyInv(AR, BR, p);
  reverse_copy(A, A + n, AR);
  polyMul(AR, p, BR, p, tmp);
  reverse(tmp, tmp + p);
  copy(tmp, tmp + p, D);
  fill(D + p, D + n, 0);
  copy(B, B + m, BR);
  polyMul(tmp, p, BR, m, AR);
  for (int i = 0; i < m; i += 1) R[i] = (A[i] - AR[i] + mod) % mod;
  fill(R + m, R + n, 0);
}
inline void drv(int x[], int y[], int n) {
  for (int i = 1; i < n; i++) y[i - 1] = mul(i, x[i]);
  y[n - 1] = 0;
}
inline void itg(int x[], int y[], int n) {
  for (int i = n - 1; i; i--) y[i] = mul(x[i - 1], qpow(i, mod - 2));
  y[0] = 0;
}
void polyLog(int A[], int B[], int deg) {
  static int tmp[maxm];
  polyInv(A, B, deg);
  drv(A, tmp, deg);
  polyMul(tmp, deg, B, deg, B);
  itg(B, B, deg);
}
void polyExp(int A[], int B[], int deg) {
  static int t[maxm];
  fill(B, B + (deg << 1), 0);
  B[0] = 1;
  for (int i = 4, j = 2; j <= deg; j = i, i <<= 1) {
    polyLog(B, t, j);
    t[0] = sub(A[0] + 1, t[0]);
    for (int k = 1; k < j; k += 1) t[k] = sub(A[k], t[k]);
    polyMul(t, j, B, j, B);
  }
}
}  // namespace FFT
using FFT::I2;
int f[maxn] = {1, 6, 1}, g[maxn], h[maxn] = {1, 1}, l[maxn];
int main() {
  FFT::init();
  int n = 1 << 15, a = 0, b = 0;
  FFT::polySqrt(f, g, n);
  scanf("%d%d", &a, &b);
  for (int i = 0; i < n; i++) h[i] = mul(I2, add(h[i], g[i]));
  FFT::polyLog(h, l, n);
  for (int i = 0; i < n; i += 1) l[i] = mul(a + 1, l[i]);
  FFT::polyExp(l, h, n);
  FFT::polyInv(g, l, n);
  FFT::polyMul(h, n, l, n, g);
  for (int i = 1; i <= b; i++) printf("%d\n", i <= a ? g[i] : 0);
  return 0;
}