#include <bits/stdc++.h>
using namespace std;

const int mod = 1000000007;

int dp[3010][2];
int ndp[3010][2];

void add(int &x, int y) {
	x += y;
	if (x >= mod) x -= mod;
}

int main () {
	int n, m; cin >> n >> m;
	for (int i = 0; i <= n; i++) {
		dp[i][i == 0] = 1;
	}
	for (int i = 0; i < m; i++) {
		memset(ndp, 0, sizeof(ndp));
		for (int j = 0; j <= n; j++) {
			for (int p = 0; p < 2; p++) {
				if (!dp[j][p]) continue;
				if (j) add(ndp[j][p | (j == 1)], dp[j][p]);
				if (j) add(ndp[j - 1][p | (j == 1)], dp[j][p]);
				if (j < n) add(ndp[j][p], dp[j][p]);
				if (j < n) add(ndp[j + 1][p], dp[j][p]);
			}
		}
		swap(ndp, dp);
	}
	int ans = 0;
	for (int i = 0; i <= n; i++) {
		add(ans, dp[i][1]);
	}
	printf("%d\n", ans);
	return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

const int mod = 1000000007;

int dp[3010][2];
int ndp[3010][2];

void add(int &x, int y) {
	x += y;
	if (x >= mod) x -= mod;
}

int main () {
	int n, m; cin >> n >> m;
	for (int i = 0; i <= n; i++) {
		dp[i][i == 0] = 1;
	}
	for (int i = 0; i < m; i++) {
		memset(ndp, 0, sizeof(ndp));
		for (int j = 0; j <= n; j++) {
			for (int p = 0; p < 2; p++) {
				if (!dp[j][p]) continue;
				if (j) add(ndp[j][p | (j == 1)], dp[j][p]);
				if (j) add(ndp[j - 1][p | (j == 1)], dp[j][p]);
				if (j < n) add(ndp[j][p], dp[j][p]);
				if (j < n) add(ndp[j + 1][p], dp[j][p]);
			}
		}
		swap(ndp, dp);
	}
	int ans = 0;
	for (int i = 0; i <= n; i++) {
		add(ans, dp[i][1]);
	}
	printf("%d\n", ans);
	return 0;
}

```