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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long n, m;
    cin >> n >> m;
    long long dp[2 * m][n + 1];
    vector<long long int> pre(n + 1, 0);
    for (long long int i = 1; i < n + 1; i++) {
      dp[0][i] = 1;
      if (i == 1)
        pre[i] = dp[0][i];
      else
        pre[i] = (pre[i - 1] + dp[0][i]) % 1000000007;
    }
    for (long long int i = 1; i < 2 * m; i++) {
      for (long long int j = 1; j < n + 1; j++) {
        dp[i][j] = pre[j] % 1000000007;
        if (j == 1)
          pre[j] = dp[i][j];
        else
          pre[j] = (pre[j - 1] + dp[i][j]) % 1000000007;
      }
    }
    long long ans = 0;
    for (long long int i = 1; i < n + 1; i++) {
      ans += dp[2 * m - 1][i];
      ans %= 1000000007;
    }
    cout << ans;
  }
  return 0;
}
