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 mod = 1e9 + 7;
signed main() {
  long long n, m;
  cin >> n >> m;
  vector<vector<long long>> ways(11, vector<long long>(1011, 0));
  for (long long i = 1; i <= 10; i++) ways[i][1] = 1;
  for (long long i = 1; i <= 1010; i++) ways[0][i] = 1, ways[1][i] = i;
  for (long long i = 2; i <= 10; i++) {
    for (long long j = 2; j <= 1010; j++) {
      ways[i][j] = (ways[i - 1][j] + ways[i][j - 1]) % mod;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++)
    (ans += (ways[m][i] * ways[m - 1][n - i + 1]) % mod) %= mod;
  cout << ans << endl;
}
