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 n, m;
const int mod = 1e9 + 7;
int64_t dp[1005][25];
int64_t f(int i, int j) {
  if (j == (2 * m)) return 1;
  if (i > n) return 0;
  if (dp[i][j] != -1) return dp[i][j];
  int64_t ans = f(i + 1, j);
  ans = (ans + f(i, j + 1)) % mod;
  return dp[i][j] = ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 0; i < 1005; ++i) {
    for (int j = 0; j < 25; ++j) {
      dp[i][j] = -1;
    }
  }
  cin >> n >> m;
  cout << f(1, 0) << endl;
  return 0;
}
