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 mod = 1e9 + 7;
void solve() {
  int n, m;
  cin >> n >> m;
  long long a[m + 1][n + 1], b[m + 1][n + 1];
  for (int i = 1; i <= n; i++) {
    a[1][i] = 1;
    b[1][i] = n - i + 1;
  }
  for (int i = 2; i <= m; i++) {
    long long sum = 0;
    for (int j = 1; j <= n; j++) {
      sum = (sum + a[i - 1][j]) % mod;
      a[i][j] = sum % mod;
    }
  }
  for (int i = 2; i <= m; i++) {
    long long sum = 0;
    for (int j = n; j >= 1; j--) {
      sum = (sum + b[i - 1][j]) % mod;
      b[i][j] = sum % mod;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = (ans + (a[m][i] * b[m][i]) % mod) % mod;
  }
  cout << ans << endl;
}
int main() { solve(); }
