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 N = 3007;
const int MOD = 1e9 + 7;

inline int add(int &a, int &b) {
	int ret = a + b;
	if(ret >= MOD) ret -= MOD;
	return ret;
}

int dp[N][N][2];

int main() {
	int n, m;
	cin >> n >> m;

	for(int i=0; i<=n; ++i) dp[m][i][1] = 1;
	for(int i=m-1; i>=0; --i) {
		for(int j=0; j<=n; ++j) {
			for(int k=0; k<2; ++k) {
				if(j > 0) {
					// RR
					dp[i][j][k] = add(dp[i][j][k], dp[i+1][j-1][k|(j==1)]);
					// RB
					dp[i][j][k] = add(dp[i][j][k], dp[i+1][j][k|(j==1)]);
				}
				if(j < n) {
					// BB
					dp[i][j][k] = add(dp[i][j][k], dp[i+1][j+1][k|(j==0)]);
					// BR
					dp[i][j][k] = add(dp[i][j][k], dp[i+1][j][k|(j==0)]);
				}
			}
		}
	}

	int res = 0;
	for(int i=0; i<=n; ++i) res = add(res, dp[0][i][0]);
	cout << res << "\n";

	return 0;
}