You are given two integers n and m. Calculate the number of pairs of arrays (a, b) such that:

  * the length of both arrays is equal to m; 
  * each element of each array is an integer between 1 and n (inclusive); 
  * a_i ≤ b_i for any index i from 1 to m; 
  * array a is sorted in non-descending order; 
  * array b is sorted in non-ascending order. 



As the result can be very large, you should print it modulo 10^9+7.

Input

The only line contains two integers n and m (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10).

Output

Print one integer – the number of arrays a and b satisfying the conditions described above modulo 10^9+7.

Examples

Input


2 2


Output


5


Input


10 1


Output


55


Input


723 9


Output


157557417

Note

In the first test there are 5 suitable arrays: 

  * a = [1, 1], b = [2, 2]; 
  * a = [1, 2], b = [2, 2]; 
  * a = [2, 2], b = [2, 2]; 
  * a = [1, 1], b = [2, 1]; 
  * a = [1, 1], b = [1, 1]. 
#include <bits/stdc++.h>
using namespace std;
long long int f[1001][1001];
int main() {
  long long int n, m, mod = 1000000007, ans = 0, k = 0;
  scanf("%lld%lld", &n, &m);
  for (long long int i = 1; i <= n || i <= 2 * m; i++) {
    for (long long int j = 0; j <= i; j++) {
      if (j == 0)
        f[j][i] = 1;
      else if (j == 1)
        f[j][i] = i;
      else {
        f[j][i] = f[j][i - 1] + f[j - 1][i - 1];
        f[j][i] %= mod;
      }
    }
  }
  for (long long int i = 1; i <= 2 * m && i <= n; i++)
    ans += f[i][n] * f[2 * m - i][2 * m - 1], ans %= mod;
  printf("%lld", ans);
}
