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 power(long long x, long long y, long long m = 1000000007) {
  if (y == 0) return 1;
  long long z = power(x, y / 2);
  z = (z * z) % m;
  if (y & 1)
    return (z * x) % m;
  else
    return (z) % m;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<long long> fact(n + 21, 0);
  fact[0] = 1;
  for (int i = 1; i <= n + 20; i++) {
    fact[i] = (i * fact[i - 1]) % 1000000007;
  }
  long long res = fact[2 * m + n - 1] % 1000000007;
  res = (res * power(fact[2 * m], 1000000007 - 2)) % 1000000007;
  cerr << res << "\n";
  res = (res * power(fact[n - 1], 1000000007 - 2)) % 1000000007;
  cout << res;
  return 0;
}
