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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class G {

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;
	boolean eof;

	static final int P = 998244353;
	static final long P2 = (long) P * P;
	static final int[] A = { 1, 1 };
	static final int[] B = { 0, 1 };

	int[] prod(int[] a, int[] b, int len) {
		int[] c = new int[Math.min(a.length + b.length - 1, len)];

		for (int k = 0; k < c.length; k++) {
			long sum = 0;
			for (int i = 0; i <= k && i < a.length; i++) {
				if (k - i < b.length) {
					sum += (long) a[i] * b[k - i];
					if (sum >= P2) {
						sum -= P2;
					}
				}
			}
			c[k] = (int) (sum % P);
		}

		return c;
	}

	int[] sum(int[] a, int[] b, int len) {
		int[] c = new int[Math.min(Math.max(a.length, b.length), len)];
		for (int i = 0; i < c.length; i++) {
			c[i] = (i < a.length ? a[i] : 0) + (i < b.length ? b[i] : 0);
			if (c[i] >= P) {
				c[i] -= P;
			}
		}
		return c;
	}

	int[] pointwise(int[] a, int[] b) {
		if (a.length != b.length) {
			throw new AssertionError();
		}
		int[] c = new int[a.length];
		for (int i = 0; i < c.length; i++) {
			c[i] = (int) ((long) a[i] * b[i] % P);
		}
		return c;
	}

	int[][] get(int n, int len) {

		if (n == 0) {
			return new int[][] { {}, { 1 } };
		}

		if ((n & 1) == 1) {
			int[][] tmp = get(n - 1, len);
			return new int[][] { sum(prod(A, tmp[0], len), tmp[1], len),
					prod(B, tmp[0], len) };
		} else {
			int[][] tmp = get(n >> 1, len);

			int needLen = FFTDiscrete.nextPowerOf2(2 * Math.max(tmp[0].length,
					tmp[1].length) - 1);

			int[] fp = Arrays.copyOf(tmp[0], needLen);
			int[] fq = Arrays.copyOf(tmp[1], needLen);

			FFTDiscrete.fft(fp, false);
			FFTDiscrete.fft(fq, false);

			int[] pp = pointwise(fp, fp);
			int[] pq = pointwise(fp, fq);
			int[] qq = pointwise(fq, fq);

			FFTDiscrete.fft(pp, true);
			FFTDiscrete.fft(pq, true);
			FFTDiscrete.fft(qq, true);

			return new int[][] {
					sum(prod(A, pp, len), prod(new int[] { 2 }, pq, len), len),
					sum(prod(B, pp, len), qq, len) };
		}
	}

	void solve() throws IOException {
		int n = nextInt();
		int k = nextInt();

		int[][] ps = get(n + 1, k + 1);
		for (int i = 1; i <= k; i++) {
			out.print((i >= ps[0].length ? 0 : ps[0][i]) + " ");
		}
		out.println();
	}

	static class FFTDiscrete {

		static final int P = 998244353;
		static final int R = 3;
		static final int INV_R = pow(R, P - 2);

		static int pow(int a, int b) {
			int ret = 1;
			for (; b > 0; b >>= 1) {
				if ((b & 1) == 1) {
					ret = (int) ((long) ret * a % P);
				}
				a = (int) ((long) a * a % P);
			}
			return ret;
		}

		public static int[] mul(int[] a, int[] b) {
			// naive multiplication when a/b is short?
			int cLen = a.length + b.length - 1;
			int len = nextPowerOf2(cLen);
			int[] da = Arrays.copyOf(a, len);
			int[] db = Arrays.copyOf(b, len);
			fft(da, false);
			fft(db, false);
			int[] dc = pointwiseMultiply(da, db);
			fft(dc, true);
			return Arrays.copyOf(dc, cLen);
		}

		public static int[] pointwiseMultiply(int[] da, int[] db) {
			int[] dc = new int[da.length];
			for (int i = 0; i < dc.length; ++i) {
				dc[i] = (int) ((long) da[i] * db[i] % P);
			}
			return dc;
		}

		static class RevOrderCache {
			private RevOrderCache() {
			}

			int[] getRevOrder(int size) {
				int log = Integer.numberOfTrailingZeros(size);
				if (log >= cache.length) {
					cache = Arrays.copyOf(cache, log + 1);
				}
				if (cache[log] == null) {
					cache[log] = makeRevOrder(log);
				}
				return cache[log];
			}

			private int[][] cache = new int[0][];
			static RevOrderCache INSTANCE = new RevOrderCache();

			private int[] makeRevOrder(int log) {
				int[] ret = new int[1 << log];
				for (int i = 0; i < (1 << log); i++) {
					ret[i] = ret[i >> 1] >> 1;
					if ((i & 1) == 1) {
						ret[i] |= 1 << (log - 1);
					}
				}
				return ret;
			}
		}

		static boolean isPowerOf2(int x) {
			return (x & (x - 1)) == 0;
		}

		static int nextPowerOf2(int x) {
			return isPowerOf2(x) ? x : Integer.highestOneBit(x) << 1;
		}

		public static void fft(int[] arr, boolean inv) {
			if ((arr.length & (arr.length - 1)) != 0) {
				throw new AssertionError();
			}
			int n = arr.length;
			int[] rev = RevOrderCache.INSTANCE.getRevOrder(n);
			for (int i = 0; i < n; ++i) {
				int j = rev[i];
				if (j > i) {
					int tmp = arr[i];
					arr[i] = arr[j];
					arr[j] = tmp;
				}
			}
			for (int p2 = 1; p2 < n; p2 <<= 1) {
				int w = pow(inv ? INV_R : R, (P - 1) / (p2 << 1));
				int[] c = new int[p2];
				c[0] = 1;
				for (int i = 1; i < p2; i++) {
					c[i] = (int) ((long) c[i - 1] * w % P);
				}
				for (int big = 0; big < n; big += p2 << 1) {
					for (int small = big; small < big + p2; ++small) {
						int i = small;
						int j = small + p2;
						int u = arr[i];
						int v = (int) ((long) c[small - big] * arr[j] % P);
						arr[i] = u + v;
						if (arr[i] >= P) {
							arr[i] -= P;
						}
						arr[j] = u - v;
						if (arr[j] < 0) {
							arr[j] += P;
						}
					}
				}
			}
			if (inv) {
				int invN = pow(n, P - 2);
				for (int i = 0; i < arr.length; ++i) {
					arr[i] = (int) ((long) arr[i] * invN % P);
				}
			}
		}
	}

	G() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new G();
	}

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return null;
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			eof = true;
			return null;
		}
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
}