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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  vector<vector<long long> > dp(2 * m + 1, vector<long long>(n + 1, 0));
  for (long long j = 1; j <= n; j++) dp[1][j] = 1;
  for (long long i = 1; i <= 2 * m; i++) dp[i][1] = 1;
  for (long long i = 2; i <= 2 * m; i++) {
    for (long long j = 2; j <= n; j++) {
      dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % 1000000007;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) ans = (ans + dp[2 * m][i]) % 1000000007;
  cout << ans << "\n";
  return 0;
}
