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;
using ll = long long;
ll f[1000001];
ll power(ll a, ll n) {
  ll res = 1;
  while (n) {
    if (n % 2 == 1) {
      res = (res % 1000000007) * (a % 1000000007) % 1000000007;
      n--;
    } else {
      a = (a % 1000000007) * (a % 1000000007) % 1000000007;
      n = n / 2;
    }
  }
  return res;
}
ll C(ll n, ll k) {
  if (k > n) return 0;
  ll res = f[n];
  res = (res * power(f[k], 1000000007 - 2)) % 1000000007;
  res = (res * power(f[n - k], 1000000007 - 2)) % 1000000007;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  f[0] = f[1] = 1;
  for (ll i = 1; i <= 1000000; i++) {
    f[i] = (f[i - 1] * i) % 1000000007;
  }
  ll n, m;
  cin >> n >> m;
  cout << C(2 * m + n - 1, n - 1) << endl;
}
