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.
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Arrays;

public class Test {
  static PrintWriter writer =
      new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
  int n, k;
  int N = 1 << 16;
  int[][] f = new int[5][N];
  int idx = 0;

  public static void main(String[] args) {
    Test te = new Test();
    te.start();
    writer.flush();
  }

  static int readInt() {
    int ans = 0;
    boolean neg = false;
    try {
      boolean start = false;
      for (int c = 0; (c = System.in.read()) != -1; ) {
        if (c == '-') {
          start = true;
          neg = true;
          continue;
        } else if (c >= '0' && c <= '9') {
          start = true;
          ans = ans * 10 + c - '0';
        } else if (start) break;
      }
    } catch (IOException e) {
    }
    return neg ? -ans : ans;
  }

  static long readLong() {
    long ans = 0;
    boolean neg = false;
    try {
      boolean start = false;
      for (int c = 0; (c = System.in.read()) != -1; ) {
        if (c == '-') {
          start = true;
          neg = true;
          continue;
        } else if (c >= '0' && c <= '9') {
          start = true;
          ans = ans * 10 + c - '0';
        } else if (start) break;
      }
    } catch (IOException e) {
    }
    return neg ? -ans : ans;
  }

  static String readString() {
    StringBuilder b = new StringBuilder();
    try {
      boolean start = false;
      for (int c = 0; (c = System.in.read()) != -1; ) {
        if (Character.isLetterOrDigit(c)) {
          start = true;
          b.append((char) c);
        } else if (start) break;
      }
    } catch (IOException e) {
    }
    return b.toString();
  }

  static int readChars(char[] a, int off) {
    int cnt = 0;
    try {
      boolean start = false;
      for (int c = 0; (c = System.in.read()) != -1; ) {
        if (Character.isLetterOrDigit(c)) {
          start = true;
          a[off + cnt++] = (char) c;
        } else if (start) break;
      }
    } catch (IOException e) {
    }
    return cnt;
  }

  void solve(int n) {
    if (n == 1) {
      f[0][0] = 1;
      f[0][1] = 0;
      f[1][0] = 1;
      f[1][1] = 1;
      idx = 1;
      return;
    }
    if (n == 2) {
      f[1][0] = 1;
      f[1][1] = 1;
      f[2][0] = 1;
      f[2][1] = 3;
      f[2][2] = 1;
      idx = 2;
      return;
    }
    if (n % 2 == 1) {
      solve(n - 1);
      int nidx = (idx + 1) % 3, pidx = (idx + 3 - 1) % 3;
      f[nidx][0] = 1;
      for (int i = 1; i <= k; i++)
        f[nidx][i] = Ntt.add(Ntt.add(f[idx][i], f[idx][i - 1]), f[pidx][i - 1]);
      idx = nidx;
      return;
    }

    solve(n / 2);
    int nidx = (idx + 1) % 3, pidx = (idx + 3 - 1) % 3, ppidx = 3;
    for (int i = 0; i + 1 <= k; i++)
      f[ppidx][i] = Ntt.add(f[idx][i + 1], Ntt.M - Ntt.add(f[pidx][i], f[pidx][i + 1]));

    Arrays.fill(f[idx], k + 1, N, 0);
    Arrays.fill(f[pidx], k + 1, N, 0);
    Arrays.fill(f[ppidx], k + 1, N, 0);

    Ntt.fft(f[idx], 0, N);
    Ntt.fft(f[pidx], 0, N);
    Ntt.fft(f[ppidx], 0, N);

    for (int i = 0; i < N; i++) f[nidx][i] = Ntt.mul(f[idx][i], f[idx][i]);
    Ntt.ifft(f[nidx], 0, N);

    for (int i = 0; i < N; i++) f[4][i] = Ntt.mul(f[pidx][i], f[pidx][i]);
    Ntt.ifft(f[4], 0, N);
    for (int i = 1; i <= k; i++) f[nidx][i] = Ntt.add(f[nidx][i], f[4][i - 1]);

    for (int i = 0; i < N; i++) f[idx][i] = Ntt.mul(f[idx][i], f[pidx][i]);
    Ntt.ifft(f[idx], 0, N);

    for (int i = 0; i < N; i++) f[pidx][i] = Ntt.mul(f[pidx][i], f[ppidx][i]);
    Ntt.ifft(f[pidx], 0, N);
    for (int i = 1; i <= k; i++) f[idx][i] = Ntt.add(f[idx][i], f[pidx][i - 1]);

    idx = nidx;
  }

  void start() {
    n = readInt();
    k = readInt();
    solve(n);
    for (int i = 1; i <= k; i++) writer.print(f[idx][i] + " ");
  }

  static class Ntt {
    static final int pri = 3;
    static final int M = 998_244_353;
    static int[] xy = new int[2];

    static int pmod(int b, int e) {
      int i = 1 << 31;
      int r = 1;
      while (i != 0) {
        r = mul(r, r);
        if ((e & i) != 0) r = mul(r, b);
        i >>>= 1;
      }
      return r;
    }

    static void fft(int[] a, int off, int len) {
      for (int i = 0; i < len; i++) {
        int j = i, x = 0, y = len - 1;
        while (y > 0) {
          x = (x << 1) + (j & 1);
          j >>>= 1;
          y >>>= 1;
        }
        if (i < x) {
          int t = a[off + i];
          a[off + i] = a[off + x];
          a[off + x] = t;
        }
      }
      for (int i = 1; i < len; i *= 2) {
        int s = pmod(pri, (M - 1) / 2 / i);
        for (int j = 0; j < len; j += 2 * i) {
          int w = 1;
          for (int k = 0; k < i; k++) {
            int u = a[off + j + k], v = mul(a[off + j + k + i], w);
            a[off + j + k] = add(u, v);
            a[off + j + k + i] = add(u, M - v);
            w = mul(w, s);
          }
        }
      }
    }

    static void ifft(int[] a, int off, int len) {
      int ll = 1, rr = len - 1;
      while (ll < rr) {
        int t = a[off + ll];
        a[off + (ll++)] = a[off + rr];
        a[off + (rr--)] = t;
      }
      fft(a, off, len);
      int inv = pmod(len, M - 2);
      for (int i = 0; i < len; i++) a[off + i] = mul(inv, a[off + i]);
    }

    static int add(int a, int b) {
      return (a + b) % M;
    }

    static int mul(int a, int b) {
      return (int) ((a * (long) b) % M);
    }

    static int inverse(int a, int b) {
      if (b == 0) {
        xy[0] = 1;
        xy[1] = 0;
        return a;
      }
      int d = inverse(b, a % b);
      swap(xy, 0, 1);
      xy[1] -= a / b * xy[0];
      return d;
    }

    static void swap(int[] a, int i, int j) {
      int t = a[i];
      a[i] = a[j];
      a[j] = t;
    }
  }
}
