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;
int m, n;
map<pair<int, int>, int> dp;
int getans(int remb, int curr) {
  if (curr == n || remb == 0) return 1;
  int ret = 0;
  if (dp.find(make_pair(remb, curr)) != dp.end())
    return dp[make_pair(remb, curr)];
  if (remb == 1) {
    return n - curr + 1;
  }
  ret = (getans(remb - 1, curr) + getans(remb, curr + 1)) % 1000000007;
  dp[make_pair(remb, curr)] = ret;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  ;
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    dp.clear();
    cin >> n >> m;
    cout << getans(2 * m, 1);
  }
  return 0;
}
