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;
vector<int> adj[1000005];
long long mod = 1e9 + 7;
int main() {
  long long int n, m;
  cin >> n >> m;
  long long d[m + 5][n + 5];
  long long d1[n + 1];
  long long d2[n + 1];
  memset(d, 0, sizeof(d));
  long long sum = 0;
  for (int i = n; i >= 1; i--) {
    memset(d, 0, sizeof(d));
    for (int j = 1; j < m; j++) {
      for (int k = 1; k <= i; k++) {
        if (j == 1) {
          d[j][k] = k;
          continue;
        }
        d[j][k] = (d[j][k] + d[j - 1][k]) % mod;
        d[j][k] = (d[j][k] + d[j][k - 1]) % mod;
      }
    }
    if (m == 1) {
      d1[i] = 1;
      d2[n - i + 1] = n - i + 1;
    } else {
      d1[i] = d[m - 1][i];
      for (int k = 1; k <= n; k++) {
        d[m][k] = (d[m][k] + d[m - 1][k]) % mod;
        d[m][k] = (d[m][k] + d[m][k - 1]) % mod;
      }
      d2[i] = d[m][n];
    }
  }
  for (int i = n; i >= 1; i--) {
    long long s1 = (d1[i] * d2[n - i + 1]) % mod;
    sum = (sum + s1) % mod;
  }
  cout << sum;
}
