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;
const int mod = 1000000007;
long long int binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t, n, k, a, b, m;
  cin >> n >> m;
  vector<long long int> fact(3001, 1), fact_inverse(3001, 1);
  for (int i = 1; i < 3001; i++) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= mod;
  }
  for (int i = 0; i < 3001; i++) {
    fact_inverse[i] = binpow(fact[i], mod - 2, mod);
  }
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    a = (((fact[i + m - 2] * fact_inverse[i - 1]) % mod) *
         fact_inverse[m - 1]) %
        mod;
    b = (((fact[n + m - i] * fact_inverse[n - i]) % mod) * fact_inverse[m]) %
        mod;
    ans += (a * b) % mod;
  }
  ans %= mod;
  cout << ans << "\n";
  return 0;
}
