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=1e9+7;
const int N=3e3+10;
int n,m,f[2][N<<1][2];
int main(){
        cin>>n>>m;
        for(int i=0;i<n;i++) f[0][i][(i==0)]=4;
        m--;
        for(int i=0;i<m;i++){
                int cur=i&1;
                memset(f[cur^1],0,sizeof(f[cur^1]));
                for(int alr=0;alr<=1;alr++)
                        for(int use=0;use<n;use++){
                                for(int add=-1;add<=1;add++) if (use+add>=0&&use+add<n){
                                        int pos=use+add;
                                        int talr=(pos==0||alr);
                                        f[cur^1][pos][talr]=(f[cur^1][pos][talr]+1LL*(1+(add==0))*f[cur][use][alr])%MOD;
                                }
                }
         
        }
        int ans=0;
        for(int i=0;i<n;i++) 
                ans=(ans+f[m&1][i][1])%MOD;
        cout<<ans%MOD;
}
