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>
int main() {
  long long int n, m;
  scanf("%lld %lld", &n, &m);
  long long int p = 1000000007;
  long long int dp[2][11][1003];
  long long int i, j;
  for (i = 1; i <= n; i++) dp[0][1][i] = dp[1][1][i] = 1;
  for (i = 2; i <= m; i++) {
    dp[0][i][n] = 1;
    for (j = n - 1; j > 0; j--)
      dp[0][i][j] = (dp[0][i][j + 1] + dp[0][i - 1][j]) % p;
    dp[1][i][1] = 1;
    for (j = 2; j <= n; j++)
      dp[1][i][j] = (dp[1][i][j - 1] + dp[1][i - 1][j]) % p;
  }
  long long int ans = 0;
  for (i = 1; i <= n; i++)
    for (j = i; j <= n; j++) ans = (ans + dp[0][m][j] * dp[1][m][i] % p) % p;
  printf("%lld\n", ans);
  return 0;
}
