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.

Examples

Input

2 3


Output

56


Input

1000 10


Output

1048576


Input

1000 3000


Output

693347555
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef bool boolean;

const int N = 3005, M = 1e9 + 7;

int n, m;
int f[N][N][2];

inline void init() {
	scanf("%d%d", &n, &m);
}

void add(int& a, int b) {
	a += b;
	if (a > M)
		a -= M;
}

inline void solve() {
	memset(f, 0, sizeof(f));
	for (int i = 0; i <= n; i++)
		f[0][i][(i == 0)] = 1;
	for (int i = 0; i < m; i++) {
		for (int j = 0; j <= n; j++) {
			for (int c = 0; c < 2; c++) {
				if (!f[i][j][c])	continue;
				if (j) {
					add(f[i + 1][j][c | (j == 1)], f[i][j][c]);	// nj = j - 1 + 1
					add(f[i + 1][j - 1][c | (j == 1)], f[i][j][c]);
				}
				if (n - j) {
					add(f[i + 1][j][c], f[i][j][c]);
					add(f[i + 1][j + 1][c], f[i][j][c]);
				}
			}
		}
	}
	int ans = 0;
	for (int i = 0; i <= n; i++)
		add(ans, f[m][i][1]);
	printf("%d\n", ans);
}

int main() {
	init();
	solve();
	return 0;
}
