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 power(long long x, long long int y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long fab(long long n) {
  long long ans = 1;
  for (long long i = 2; i <= n; i++) {
    ans = (ans * i) % 1000000007;
  }
  return ans;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  long long upper = fab(n + 2 * m - 1),
            lower = (fab(2 * m) * fab(n - 1)) % 1000000007;
  lower = power(lower, 1000000007 - 2);
  upper = (upper * lower) % 1000000007;
  cout << upper << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t-- > 0) {
    solve();
  }
  return 0;
}
