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 mod = 1e9 + 7;
int k, mnc, mxc, fac, temp;
inline void modinv(long long int inv, long long int mod) {
  if (inv >= 0)
    inv = inv % mod;
  else
    inv = mod - ((0 - inv) % mod);
  mnc = 1, mxc = 0;
  int min = inv;
  int max = mod;
  while (min > 1) {
    fac = max / min;
    swap(mnc, mxc);
    mnc -= fac * mxc;
    max = max % min;
    swap(min, max);
  }
  if (mnc >= 0)
    k = mnc;
  else
    k = mod - ((0 - mnc) % mod);
}
int main() {
  long long int n, m;
  cin >> n >> m;
  n = 2 * m + n - 1;
  m *= 2;
  long long int ans = 1;
  for (long long int i = 0; i < m; i++) {
    modinv(i + 1, mod);
    ans = ans * (n - i) % mod * k % mod;
  }
  cout << ans << endl;
}
