#include <bits/stdc++.h>
using namespace std;
const long long MOD = (1e9)+7;

int n, m;
long long dp[3001][3001][2], ans;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m;
	dp[0][0][1] = 1;
	for(int j = 1; j <= n; j++) dp[0][j][0] = 1;
	for(int i = 1; i <= m; i++) for(int j = 0; j <= n; j++) {
		dp[i][j][0] = (j > 0 ? dp[i-1][j-1][0] : 0)+2*dp[i-1][j][0]+(j < n ? dp[i-1][j+1][0] : 0);
		dp[i][j][1] = (j > 0 ? dp[i-1][j-1][1] : 0)+2*dp[i-1][j][1]+(j < n ? dp[i-1][j+1][1] : 0);
		if(!j or j == n) {
			dp[i][j][0] -= dp[i-1][j][0];
			dp[i][j][1] -= dp[i-1][j][1];
			if(!j) {
				dp[i][j][0] = 0;
				dp[i][j][1] += (j < n ? dp[i-1][j+1][0] : 0);
			}
		}
		if(j == 1) {
			dp[i][j][0] -= dp[i-1][j][0];
			dp[i][j][1] += dp[i-1][j][0];
		}
		dp[i][j][0] = (dp[i][j][0]+MOD)%MOD;
		dp[i][j][1] = (dp[i][j][1]+MOD)%MOD;
	}
	for(int j = 0; j <= n; j++) ans = (ans+dp[m][j][1])%MOD;
	cout << ans;
}

### Prompt

Create a solution in Cpp for 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 long long MOD = (1e9)+7;

int n, m;
long long dp[3001][3001][2], ans;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m;
	dp[0][0][1] = 1;
	for(int j = 1; j <= n; j++) dp[0][j][0] = 1;
	for(int i = 1; i <= m; i++) for(int j = 0; j <= n; j++) {
		dp[i][j][0] = (j > 0 ? dp[i-1][j-1][0] : 0)+2*dp[i-1][j][0]+(j < n ? dp[i-1][j+1][0] : 0);
		dp[i][j][1] = (j > 0 ? dp[i-1][j-1][1] : 0)+2*dp[i-1][j][1]+(j < n ? dp[i-1][j+1][1] : 0);
		if(!j or j == n) {
			dp[i][j][0] -= dp[i-1][j][0];
			dp[i][j][1] -= dp[i-1][j][1];
			if(!j) {
				dp[i][j][0] = 0;
				dp[i][j][1] += (j < n ? dp[i-1][j+1][0] : 0);
			}
		}
		if(j == 1) {
			dp[i][j][0] -= dp[i-1][j][0];
			dp[i][j][1] += dp[i-1][j][0];
		}
		dp[i][j][0] = (dp[i][j][0]+MOD)%MOD;
		dp[i][j][1] = (dp[i][j][1]+MOD)%MOD;
	}
	for(int j = 0; j <= n; j++) ans = (ans+dp[m][j][1])%MOD;
	cout << ans;
}

```