Problem:
Score : 900 points

Problem Statement

Joisino has a lot of red and blue bricks and a large box.
She will build a tower of these bricks in the following manner.

First, she will pick a total of N bricks and put them into the box.
Here, there may be any number of bricks of each color in the box, as long as there are N bricks in total.
Particularly, there may be zero red bricks or zero blue bricks.
Then, she will repeat an operation M times, which consists of the following three steps:

Take out an arbitrary brick from the box.

Put one red brick and one blue brick into the box.

Take out another arbitrary brick from the box.

After the M operations, Joisino will build a tower by stacking the 2 \times M bricks removed from the box, in the order they are taken out.
She is interested in the following question: how many different sequences of colors of these 2 \times M bricks are possible?
Write a program to find the answer.
Since it can be extremely large, print the count modulo 10^9+7.

Constraints

1 \leq N \leq 3000

1 \leq M \leq 3000

Input

Input is given from Standard Input in the following format:

N M

Output

Print the count of the different possible sequences of colors of 2 \times M bricks that will be stacked, modulo 10^9+7.

Sample Input 1

2 3

Sample Output 1

56

A total of six bricks will be removed from the box. The only impossible sequences of colors of these bricks are the ones where the colors of the second, third, fourth and fifth bricks are all the same. Therefore, there are 2^6 - 2 \times 2 \times 2 = 56 possible sequences of colors.

Sample Input 2

1000 10

Sample Output 2

1048576

Sample Input 3

1000 3000

Sample Output 3

693347555

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

/*
cat <<EOF >mistaken-paste
*/

#pragma GCC diagnostic ignored "-Wincompatible-pointer-types"
#define _USE_MATH_DEFINES

#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>
#include <time.h>

#define BIG 2000000007
#define VERYBIG 2000000000000007LL

#define MOD 1000000007
#define FOD  998244353
typedef uint64_t ull;
typedef  int64_t sll;

#define N_MAX 1000000
#define M_MAX 200000

#ifdef __cplusplus
#include <queue>
#include <stack>
#include <tuple>
#include <set>
#include <map>
#include <string>

using std::queue;
using std::priority_queue;
using std::stack;
using std::tuple;
using std::set;
using std::map;
using std::vector;
using std::greater;
using std::pair;
using std::string;

#endif

typedef struct {
	int32_t a;
	int32_t b;
} hw;

typedef struct {
	sll a;
	sll b;
} hwll;

typedef struct {
	sll a;
	sll b;
	sll c;
} hwllc;

typedef struct {
	hwll a;
	hwll b;
} linell;

ull n, m;
ull h, w;
ull k;
ull q;
sll va, vb, vc, vd, ve, vf;
ull ua, ub, uc, ud, ue, uf;
long double vra, vrb, vrc;
double vda, vdb, vdc;
char ch, dh;

ull umin (ull x, ull y) {
	return (x < y) ? x : y;
}

ull umax (ull x, ull y) {
	return (x > y) ? x : y;
}

sll smin (sll x, sll y) {
	return (x < y) ? x : y;
}

sll smax (sll x, sll y) {
	return (x > y) ? x : y;
}

ull gcd (ull x, ull y) {
	if (x < y) {
		return gcd(y, x);
	} else if (y == 0) {
		return x;
	} else {
		return gcd(y, x % y);
	}
}

ull bitpow (ull a, ull x, ull modulo) {
	ull result = 1;
	while (x) {
		if (x & 1) {
			result *= a;
			result %= modulo;
		}
		x /= 2;
		a = (a * a) % modulo;
	}
	return result;
}

ull divide (ull a, ull b, ull modulo) {
	return (a * bitpow(b, modulo - 2, modulo)) % modulo;
}

ull udiff (ull a, ull b) {
	if (a >= b) {
		return a - b;
	} else {
		return b - a;
	}
}

sll sdiff (sll a, sll b) {
	if (a >= b) {
		return a - b;
	} else {
		return b - a;
	}
}

int bitcount (ull n) {
	int result = 0;
	while (n) {
		if (n & 1) result++;
		n /= 2;
	}
	return result;
}

// double distance (sll x1, sll y1, sll x2, sll y2) {
// 	double xdist2, ydist2, origindist, dist;

// 	xdist2 = (x1 - x2) * (x1 - x2);
// 	ydist2 = (y1 - y2) * (y1 - y2);
// 	return sqrt(xdist2 + ydist2);
// }

int32_t pullcomp (const void *left, const void *right) {
	ull l = *(ull*)left;
	ull r = *(ull*)right;
	if (l < r) {
		return -1;
	}
	if (l > r) {
		return +1;
	}
	return 0;
}

int32_t psllcomp (const void *left, const void *right) {
	sll l = *(sll*)left;
	sll r = *(sll*)right;
	if (l < r) {
		return -1;
	}
	if (l > r) {
		return +1;
	}
	return 0;
}

int32_t pcharcomp (const void *left, const void *right) {
	char l = *(char*)left;
	char r = *(char*)right;
	if (l < r) {
		return -1;
	}
	if (l > r) {
		return +1;
	}
	return 0;
}

int32_t pdoublecomp (const void *left, const void *right) {
	double l = *(double*)left;
	double r = *(double*)right;
	if (l < r) {
		return -1;
	}
	if (l > r) {
		return +1;
	}
	return 0;
}

int32_t pstrcomp (const void *left, const void *right) {
	char* l = *(char**)left;
	char* r = *(char**)right;

	return strcmp(l, r);
}

int32_t phwllABcomp (const void *left, const void *right) {
	hwll l = *(hwll*)left;
	hwll r = *(hwll*)right;
	if (l.a < r.a) {
		return -1;
	}
	if (l.a > r.a) {
		return +1;
	}
	if (l.b < r.b) {
		return -1;
	}
	if (l.b > r.b) {
		return +1;
	}
	return 0;
}

int32_t phwllREVcomp (const void *left, const void *right) {
	hwll l = *(hwll*)left;
	hwll r = *(hwll*)right;
	if (l.b < r.b) {
		return -1;
	}
	if (l.b > r.b) {
		return +1;
	}
	if (l.a < r.a) {
		return -1;
	}
	if (l.a > r.a) {
		return +1;
	}
	return 0;
}

int32_t ptriplecomp (const void *left, const void *right) {
	hwllc l = *(hwllc*)left;
	hwllc r = *(hwllc*)right;

	if (l.a < r.a) {
		return -1;
	}
	if (l.a > r.a) {
		return +1;
	}
	if (l.b < r.b) {
		return -1;
	}
	if (l.b > r.b) {
		return +1;
	}
	if (l.c < r.c) {
		return -1;
	}
	if (l.c > r.c) {
		return +1;
	}
	return 0;
}

bool isinrange (sll left, sll x, sll right) {
	return (left <= x && x <= right);
}

bool isinrange_soft (sll left, sll x, sll right) {
	return (left <= x && x <= right) || (left >= x && x >= right);
}

sll a[N_MAX];
// sll a[3001][3001];
sll b[N_MAX];
// sll b[3001][3001];
sll c[N_MAX];
sll d[N_MAX];
sll e[N_MAX];
char s[N_MAX + 1];
// char s[3010][3010];
char t[N_MAX + 1];
// char t[3010][3010];
// hwll xy[N_MAX];
// sll table[1000][1000];

ull dp[3001][3001][2];

ull solve () {
	sll i, j, ki, li;
	ull result = 0;
	// sll result = 0;
	double dresult = 0;
	// ull maybe = 0;
	sll maybe = 0;
	// ull sum = 0;
	sll sum = 0;
	sll item;
	ull *dpcell;

	for (i = 0; i <= m; i++) {
		for (j = 0; j <= n; j++) {
			for (ki = 0; ki < 2; ki++) {
				if (!i) {
					dp[i][j][ki] = (j ? 1 : 0) ^ (ki ? 1 : 0);
					continue;
				}

				dp[i][j][ki] = 0;
				if (j > 0) {
					if (j == 1) {
						if (ki) {
							dp[i][j][ki] += (dp[i - 1][j][0] + dp[i - 1][j][1]);
						}
					} else {
						dp[i][j][ki] += dp[i - 1][j][ki];
					}

					dp[i][j][ki] += dp[i - 1][j - 1][ki];
				}
				if (j < n) {
					dp[i][j][ki] += dp[i - 1][j][ki];

					if (j == 0) {
						if (ki) {
							dp[i][j][ki] += (dp[i - 1][j + 1][0] + dp[i - 1][j + 1][1]);
						}
					} else {
						dp[i][j][ki] += dp[i - 1][j + 1][ki];
					}
				}

				dp[i][j][ki] %= MOD;
			}
		}
	}

	result = 0;
	for (j = 0; j <= n; j++) {
		result = (result + dp[m][j][1]) % MOD;
	}

	printf("%lld\n", result);
	// printf("%.15lf\n", dresult);
	// puts(s);

	return 0;

	success:
	// puts("YES");
	// puts("Yes");
	// printf("%llu\n", result);
	// puts("0");
	puts("AC");
	return 0;

	fail:
	// puts("NO");
	// puts("No");
	// puts("0");
	// puts("-1");
	// puts("-1 -1 -1");
	puts("WA");
	return 1;
}

int32_t main (void) {
	int32_t i, j;
	int32_t x, y;

	// scanf("%llu%llu", &h, &w);
	scanf("%llu%llu", &n, &m);
	// scanf("%llu", &k, &n, &m);
	// scanf("%lld%lld", &va, &vb, &vc, &vd);
	// scanf("%llu%llu", &ua, &ub, &uc, &ud);
	// scanf("%llu", &q);
	// scanf("%s", s);
	// scanf("%s", t);
	// for (i = 0; i < n; i++) {
	// 	// scanf("%lld%lld", &xy[i].a, &xy[i].b);
	// 	scanf("%lld", &a[i]);
	// 	// scanf("%lld", &b[i]);
	// 	// scanf("%lld", &c[i]);
	// 	// scanf("%lld", &d[i]);
	// 	// a[i]--;
	// 	// b[i]--;
	// 	// c[i]--;
	// }
	// scanf("%llu", &m, &k);
	// scanf("%llu", &q);
	// for (i = 0; i < k; i++) {
	// 	scanf("%lld", &e[i]);
	// 	e[i]--;
	// 	scanf("%lld", &d[i]);
	// }

	// for (i = 0; i < n; i++) {
	// 	for (j = 0; j < n; j++) {
	// 		scanf("%llu", &table[i][j]);
	// 	}
	// }
	for (i = 0; i < h; i++) {
		scanf("%s", &s[i]);
	}

	solve();

	return 0;
}