Find the number, modulo 998244353, of sequences of length N consisting of 0, 1 and 2 such that none of their contiguous subsequences totals to X.

Constraints

* 1 \leq N \leq 3000
* 1 \leq X \leq 2N
* N and X are integers.

Input

Input is given from Standard Input in the following format:


N X


Output

Print the number, modulo 998244353, of sequences that satisfy the condition.

Examples

Input

3 3


Output

14


Input

8 6


Output

1179


Input

10 1


Output

1024


Input

9 13


Output

18402


Input

314 159


Output

459765451
#include <bits/stdc++.h>

#define debug(...) fprintf(stderr, __VA_ARGS__)

#ifndef AT_HOME
#define getchar() IO::myGetchar()
#define putchar(x) IO::myPutchar(x)
#endif

namespace IO {
	static const int IN_BUF = 1 << 23, OUT_BUF = 1 << 23;

	inline char myGetchar() {
		static char buf[IN_BUF], *ps = buf, *pt = buf;
		if (ps == pt) {
			ps = buf, pt = buf + fread(buf, 1, IN_BUF, stdin);
		}
		return ps == pt ? EOF : *ps++;
	}

	template<typename T>
	inline bool read(T &x) {
		bool op = 0;
		char ch = getchar();
		x = 0;
		for (; !isdigit(ch) && ch != EOF; ch = getchar()) {
			op ^= (ch == '-');
		}
		if (ch == EOF) {
			return false;
		}
		for (; isdigit(ch); ch = getchar()) {
			x = x * 10 + (ch ^ '0');
		}
		if (op) {
			x = -x;
		}
		return true;
	}

	inline int readStr(char *s) {
		int n = 0;
		char ch = getchar();
		for (; isspace(ch) && ch != EOF; ch = getchar())
			;
		for (; !isspace(ch) && ch != EOF; ch = getchar()) {
			s[n++] = ch;
		}
		s[n] = '\0';
		return n;
	}

	inline void myPutchar(char x) {
		static char pbuf[OUT_BUF], *pp = pbuf;
		struct _flusher {
			~_flusher() {
				fwrite(pbuf, 1, pp - pbuf, stdout);
			}
		};
		static _flusher outputFlusher;
		if (pp == pbuf + OUT_BUF) {
			fwrite(pbuf, 1, OUT_BUF, stdout);
			pp = pbuf;
		}
		*pp++ = x;
	}

	template<typename T>
	inline void print_(T x) {
		if (x == 0) {
			putchar('0');
			return;
		}
		static int num[40];
		if (x < 0) {
			putchar('-');
			x = -x;
		}
		for (*num = 0; x; x /= 10) {
			num[++*num] = x % 10;
		}
		while (*num){
			putchar(num[*num] ^ '0');
			--*num;
		}
	}

	template<typename T>
	inline void print(T x, char ch = '\n') {
		print_(x);
		putchar(ch);
	}

	inline void printStr_(const char *s, int n = -1) {
		if (n == -1) {
			n = strlen(s);
		}
		for (int i = 0; i < n; ++i) {
			putchar(s[i]);
		}
	}

	inline void printStr(const char *s, int n = -1, char ch = '\n') {
		printStr_(s, n);
		putchar(ch);
	}
}
using namespace IO;

const int N = 3005, P = 998244353;

int n, X, C[N][N], f[N][N << 1];

int main() {
	read(n), read(X);
	for (int i = 0; i <= n; ++i) {
		C[i][0] = C[i][i] = 1;
		for (int j = 1; j < i; ++j) {
			C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
		}
	}
	f[0][0] = 1;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= 2 * i && j < X; ++j) {
			f[i][j] = (f[i - 1][j - 1] + (j > 1 ? f[i - 1][j - 2] : 0)) % P;
		}
		for (int j = X + 1; j <= 2 * i; j += 2) {
			int c = (j - X + 1) / 2;
			if (i >= 2 * c && j >= 4 * c) {
				f[i][j] = f[i - 2 * c][j - 4 * c];
			} else if (2 * (i - 2 * c) == j - 4 * c) {
				f[i][j] = 1;
			}
		}
	}
	int ans = 0;
	for (int i = 0; i <= n; ++i) {
		for (int j = 0; j <= 2 * i; ++j) {
			ans = (ans + 1ll * C[n][i] * f[i][j]) % P;
			// debug("%d ", f[i][j]);
		}
		// debug("\n");
	}
	print(ans);
}
