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;
void jakos() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
}
const long long mod = 1e9 + 7;
const long long base = 179;
const long long INF = 1e9;
const long long N = 1e5;
long long n, m;
vector<vector<long long>> c(3000, vector<long long>(3000, 0));
long long solve(long long cn, long long cm) { return c[cn + cm - 1][cm + 1]; }
signed main() {
  jakos();
  cin >> n >> m;
  c[0][1] = 1;
  for (long long i = 1; i <= 2000; i++) {
    for (long long j = 1; j <= i + 1; j++) {
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
    }
  }
  cout << solve(n, 2 * m) << '\n';
  return 0;
}
