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>
#pragma GCC optimize("Ofast")
using namespace std;
long long int fact[100005];
long long int power(long long int x, long long int y, long long int m) {
  if (y == 0) return 1;
  long long int p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long int ritik(long long int a, long long int m) {
  return power(a, m - 2, m);
}
int main() {
  fact[0] = 1;
  for (long long int i = 1; i < 100005; i++)
    fact[i] = (fact[i - 1] * i) % 1000000007;
  long long int n, m;
  cin >> n >> m;
  long long int ans =
      (fact[n + 2 * m - 1] *
       ((ritik(fact[2 * m], 1000000007) * ritik(fact[n - 1], 1000000007)) %
        1000000007)) %
      1000000007;
  cout << ans << '\n';
}
