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

const int MAXN = 3005, mod = 1e9+7;

int n, m;

long long dp[MAXN][MAXN][2];
int power[MAXN];

int main()
{
	cin >> n >> m;
	power[0] = 1;
	for (int i = 1; i <= m; i++) power[i] = power[i-1]*4ll%mod;
	dp[0][0][0] = 1;
	for (int i = 0; i < m; i++) 
		for (int j = 0; j <= n; j++) {
			for (int k = 0; k < 2; k++) { 
				if (j != n) (dp[i+1][j+1][k||(j+1==n)] += dp[i][j][k]) %= mod;
				if (j) (dp[i+1][j-1][k] += dp[i][j][k]) %= mod;
				if (k == 1 || j == n) 
					(dp[i+1][j][1] += dp[i][j][k]*((j!=0)+(j!=n))) %= mod;
				else if (j == n-1) {
					(dp[i+1][j][1] += dp[i][j][k]) %= mod;
					(dp[i+1][j][0] += dp[i][j][k]*(j!=0)) %= mod;
				} else (dp[i+1][j][k] += dp[i][j][k]*((j!=0)+(j!=n))) %= mod;
			}
		}
	int cur = power[m];
	for (int i = 1; i < m; i++) {
		for (int j = 0; j <= n; j++)
			cur = ((cur-4ll*dp[i][j][1]*power[m-i-1])%mod+mod)%mod;
	}
	cout << cur << endl;
	return 0;
}
