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;
int n, m;
long long f1[1010][1010], f2[1010][1010];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    f1[1][i] = f2[1][i] = 1;
  }
  for (int i = 2; i <= m; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = 1; k <= j; k++) {
        f1[i][j] = (f1[i][j] + f1[i - 1][k]) % 1000000007;
      }
      for (int k = n; k >= j; k--) {
        f2[i][j] = (f2[i][j] + f2[i - 1][k]) % 1000000007;
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      ans = (ans + f1[m][i] * f2[m][j]) % 1000000007;
    }
  }
  printf("%d\n", ans);
  return 0;
}
