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 solve() {
  long long int dp[1005][13] = {0};
  for (int i = 1; i <= 1000; i++) dp[i][1] = i;
  for (int i = 1; i <= 1000; i++) {
    for (int j = 2; j <= 10; j++)
      dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % 1000000007;
  }
  long long int n, m;
  cin >> n >> m;
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += (dp[i][m] * dp[n - i + 1][m] - dp[i - 1][m] * dp[n - i + 1][m]) %
           1000000007;
    ans = (ans + 1000000007) % 1000000007;
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
}
