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 mod = 1000000007;
int N, M;
int dp[3002][3002][2];

int main() {
    scanf("%d %d", &N, &M);
    N--;
    M--;

    for(int m = 0; m <= M; m++) {
        for(int r = 0; r <= N; r++) {
            for(int z = 0; z < 2; z++) {
                if(m == 0) {
                    dp[m][r][z] = z | (r == 0);
                    continue;
                }
                dp[m][r][z] = 2 * dp[m - 1][r][ z | (r == 0) ] % mod;
                if(r != 0) dp[m][r][z] = (dp[m][r][z] + dp[m - 1][r - 1][ z | (r == 0) ]) % mod;
                if(r != N) dp[m][r][z] = (dp[m][r][z] + dp[m - 1][r + 1][ z | (r == 0) ]) % mod;
            }
        }
    }

    int ans = 0;
    for(int i = 0; i <= N; i++) {
        ans += dp[M][i][0];
        ans %= mod;
    }
    ans = 4LL * ans % mod;
    printf("%d", ans);
}
