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 minv(long long a) {
  long long tmp = 1000000007 - 2, sol = 1, pw = a;
  while (tmp) {
    if (tmp % 2) sol = (sol * pw) % 1000000007;
    pw = (pw * pw) % 1000000007;
    tmp /= 2;
  }
  return sol;
}
long long comb(long long a, long long b) {
  if (a < b) return 1;
  long long sol = 1;
  for (long long i = 1; i <= b; i++, a--) {
    sol = (((sol * a) % 1000000007) * minv(i)) % 1000000007;
  }
  return sol;
}
int main() {
  int t = 1;
  while (t--) {
    long long n, m, sol = 0;
    cin >> n >> m;
    for (int i = n; i >= 1; i--) {
      long long ar = i, br = n - i + 1;
      sol = (sol + (comb(m + ar - 2, ar - 1) * comb(br + m - 1, br - 1)) %
                       1000000007) %
            1000000007;
    }
    cout << sol << endl;
  }
  return 0;
}
