#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;

int add( int x, int y ){
	return (x + y) % mod;
}

int dp[3010][3010][2], n, m, f, t;
ll sol;

int main(){
	scanf("%d %d", &n, &m);
	for ( int i = 1; i <= n; i++ )
		dp[0][i][0] = 1;
	dp[0][0][1] = 1;
	for ( int i = 0; i < m; i++ )
		for ( int j = 0; j <= n; j++ )
			for ( int k = 0; k < 2; k++ ){
				if ( j > 0 ) dp[i + 1][j][k | (j == 1)] += dp[i][j][k];
				if ( j > 0 ) dp[i + 1][j][k | (j == 1)] %= mod;
				if ( j > 0 ) dp[i + 1][j-1][k | (j == 1)] += dp[i][j][k];
				if ( j > 0 ) dp[i + 1][j-1][k | (j == 1)] %= mod;
				if ( j < n ) dp[i + 1][j][k] += dp[i][j][k];
				if ( j < n ) dp[i + 1][j][k] %= mod;
				if ( j < n ) dp[i + 1][j+1][k] += dp[i][j][k];
				if ( j < n ) dp[i + 1][j+1][k] %= mod;
			}
	for ( int i = 0; i <= n; i++ )
		sol += dp[m][i][1];
	printf("%lld\n", sol % ll(mod));
	
	return 0;
}

### Prompt

Your task is to create a cpp solution to 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 <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;

int add( int x, int y ){
	return (x + y) % mod;
}

int dp[3010][3010][2], n, m, f, t;
ll sol;

int main(){
	scanf("%d %d", &n, &m);
	for ( int i = 1; i <= n; i++ )
		dp[0][i][0] = 1;
	dp[0][0][1] = 1;
	for ( int i = 0; i < m; i++ )
		for ( int j = 0; j <= n; j++ )
			for ( int k = 0; k < 2; k++ ){
				if ( j > 0 ) dp[i + 1][j][k | (j == 1)] += dp[i][j][k];
				if ( j > 0 ) dp[i + 1][j][k | (j == 1)] %= mod;
				if ( j > 0 ) dp[i + 1][j-1][k | (j == 1)] += dp[i][j][k];
				if ( j > 0 ) dp[i + 1][j-1][k | (j == 1)] %= mod;
				if ( j < n ) dp[i + 1][j][k] += dp[i][j][k];
				if ( j < n ) dp[i + 1][j][k] %= mod;
				if ( j < n ) dp[i + 1][j+1][k] += dp[i][j][k];
				if ( j < n ) dp[i + 1][j+1][k] %= mod;
			}
	for ( int i = 0; i <= n; i++ )
		sol += dp[m][i][1];
	printf("%lld\n", sol % ll(mod));
	
	return 0;
}

```