You are given a grid, consisting of 2 rows and n columns. Each cell of this grid should be colored either black or white.

Two cells are considered neighbours if they have a common border and share the same color. Two cells A and B belong to the same component if they are neighbours, or if there is a neighbour of A that belongs to the same component with B.

Let's call some bicoloring beautiful if it has exactly k components.

Count the number of beautiful bicolorings. The number can be big enough, so print the answer modulo 998244353.

Input

The only line contains two integers n and k (1 ≤ n ≤ 1000, 1 ≤ k ≤ 2n) — the number of columns in a grid and the number of components required.

Output

Print a single integer — the number of beautiful bicolorings modulo 998244353.

Examples

Input

3 4


Output

12


Input

4 1


Output

2


Input

1 2


Output

2

Note

One of possible bicolorings in sample 1:

<image>
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Solve6 {

    public static void main(String[] args) throws IOException {
        PrintWriter pw = new PrintWriter(System.out);
        new Solve6().solve(pw);
        pw.flush();
        pw.close();
    }

    public void solve(PrintWriter pw) throws IOException {
        FastReader sc = new FastReader();
        int n = sc.nextInt();
        k = sc.nextInt();
        dp = new long[n][k + 1][5];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= k; j++) {
                Arrays.fill(dp[i][j], -1);
            }
        }
        dp(n - 1, 0, 0);
        pw.println(dp[n - 1][0][0]);
    }
    long[][][] dp;
    int k;
    final int MOD = 998244353;

    public long dp(int j, int pre, int num) {
        if (num > k) {
            return 0;
        }
        if (j < 0) {
            return num == k ? 1 : 0;
        }
        if (dp[j][num][pre] != -1) {
            return dp[j][num][pre];
        }
        long s = 0;
        for (int i = 1; i <= 4; i++) {
            int succ = num;
            switch (pre) {
                case 0:
                    if (i == 1 || i == 4) {
                        succ = 1;
                    } else {
                        succ = 2;
                    }
                    break;
                case 1:
                    if (i != 1) {
                        ++succ;
                    }
                    break;
                case 2:
                    if (i == 3) {
                        succ += 2;
                    }
                    break;
                case 3:
                    if (i == 2) {
                        succ += 2;
                    }
                    break;
                case 4:
                    if (i != 4) {
                        ++succ;
                    }
            }
            s += dp(j - 1, i, succ) % MOD;
        }
        return dp[j][num][pre] = s % MOD;
    }

    static class FastReader {

        StringTokenizer st;
        BufferedReader br;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public String next() throws IOException {
            if (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }

        public boolean hasNext() throws IOException {
            if (st != null && st.hasMoreTokens()) {
                return true;
            }
            String s = br.readLine();
            if (s == null || s.isEmpty()) {
                return false;
            }
            st = new StringTokenizer(s);
            return true;
        }

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

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }
    }
}
