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 main() {
  long long int t = 1;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  while (t--) {
    long long int n, sum = 0, ct = 0, m, r, ans = 0, k;
    cin >> n >> m;
    long long int dp[n][2 * m];
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for (long long int i = 1; i < n; i++) dp[i][0] = i + 1;
    for (long long int j = 1; j < 2 * m; j++) dp[0][j] = 1;
    for (long long int i = 1; i < n; i++) {
      for (long long int j = 1; j < 2 * m; j++)
        dp[i][j] = (dp[i - 1][j] % 1000000007 + dp[i][j - 1] % 1000000007) %
                   1000000007;
    }
    cout << dp[n - 1][2 * m - 1] << '\n';
  }
  return 0;
}
