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 <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAX_N = 3005, MOD = 1000000007;

int f[MAX_N][2], g[MAX_N][2];
int N, M;

inline void update(int &x, int y) {
  x = (x + y) % MOD;
}

int main() {
  scanf("%d%d", &N, &M);
  for (int i = 0; i <= N; ++i) f[i][0] = 1;
  for (int i = 1; i <= M; ++i) {
    memset(g, 0, sizeof g);
    for (int j = 0; j <= N; ++j) 
      for (int x = 0; x < 2; ++x) {
	int R = j, B = N - j;
	if (R >= 1) {
	  update(g[j - 1][(R == 1) | x], f[j][x]);
	  update(g[j][(R == 1) | x], f[j][x]);
	}
	if (B >= 1) {
	  update(g[j + 1][(B == N) | x], f[j][x]);
	  update(g[j][(B == N) | x], f[j][x]);
	}
      }
    memcpy(f, g, sizeof f);
  }
  int result = 0;
  for (int i = 0; i <= N; ++i)
    update(result, f[i][1]);
  printf("%d\n", result);
  return 0;
}
