#include <iostream>
#include <cstring>

using namespace std;

const int NMAX = 3000 + 5;
const int MMAX = 3000 + 5;
const int MOD = 1000000000 + 7;

int dp[2][2 * NMAX][2];

void upd(int &where, int val) {
    where += val;
    if (where >= MOD)
        where -= MOD;
}

int main()
{
    int N, M;
    cin >> N >> M;

    M *= 2;
    int L = 2 * N + 1;

    for (int i = 1; i <= L; i += 2)
        dp[0][i][i <= 2] = 1;
    for (int i = 0; i < M; ++ i) {
        memset(dp[(i + 1) & 1], 0, sizeof(dp[0]));
        for (int j = 1; j <= L; ++ j) {
            upd(dp[(i + 1) & 1][j - 1][(j - 1 <= 2)], dp[i & 1][j][0]);
            upd(dp[(i + 1) & 1][j + 1][0], dp[i & 1][j][0]);
            upd(dp[(i + 1) & 1][j - 1][1], dp[i & 1][j][1]);
            upd(dp[(i + 1) & 1][j + 1][1], dp[i & 1][j][1]);
        }
    }

    int ans = 0;
    for (int i = 1; i <= L; ++ i)
        upd(ans, dp[M & 1][i][1]);
    cout << ans << '\n';
    return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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 <iostream>
#include <cstring>

using namespace std;

const int NMAX = 3000 + 5;
const int MMAX = 3000 + 5;
const int MOD = 1000000000 + 7;

int dp[2][2 * NMAX][2];

void upd(int &where, int val) {
    where += val;
    if (where >= MOD)
        where -= MOD;
}

int main()
{
    int N, M;
    cin >> N >> M;

    M *= 2;
    int L = 2 * N + 1;

    for (int i = 1; i <= L; i += 2)
        dp[0][i][i <= 2] = 1;
    for (int i = 0; i < M; ++ i) {
        memset(dp[(i + 1) & 1], 0, sizeof(dp[0]));
        for (int j = 1; j <= L; ++ j) {
            upd(dp[(i + 1) & 1][j - 1][(j - 1 <= 2)], dp[i & 1][j][0]);
            upd(dp[(i + 1) & 1][j + 1][0], dp[i & 1][j][0]);
            upd(dp[(i + 1) & 1][j - 1][1], dp[i & 1][j][1]);
            upd(dp[(i + 1) & 1][j + 1][1], dp[i & 1][j][1]);
        }
    }

    int ans = 0;
    for (int i = 1; i <= L; ++ i)
        upd(ans, dp[M & 1][i][1]);
    cout << ans << '\n';
    return 0;
}

```