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 int poww(long long int x, long long int n, long long int p) {
  if (x == 1 || n == 0) return 1;
  long long int ans = poww(x, n / 2, p);
  if (n & 1) {
    return (((ans * ans) % p) * (x % p)) % p;
  } else
    return ((ans * ans) % p);
}
long long int fac(long long int n) {
  if (n <= 1) return 1;
  return (n * fac(n - 1)) % 1000000007;
}
long long int com(long long int n, long long int r, long long int p) {
  long long int up = fac(n);
  long long int l1 = poww(fac(r), p - 2, p);
  long long int l2 = poww(fac(n - r), p - 2, p);
  return (((up * l1) % p) * (l2)) % p;
}
int main() {
  long long int n, m;
  cin >> n >> m;
  long long int t = n - 1;
  n = n + 2 * m - 1;
  m = t;
  cout << com(n, m, 1000000007);
  return 0;
}
