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;
long long const N = 1e3;
long long INF = 0x3f3f3f3f;
long long t, n, m, dp1[15][1050], dp2[15][1050];
long long const mod = 1e9 + 7;
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= n; j++) {
      if (j == 1)
        dp1[i][j] = 1;
      else
        dp1[i][j] = dp1[i - 1][j] + dp1[i][j - 1];
      dp1[i][j] %= mod;
    }
  }
  for (int i = 1; i <= m; i++) {
    for (int j = n; j >= 1; j--) {
      if (j == n)
        dp2[i][j] = 1;
      else
        dp2[i][j] = dp2[i - 1][j] + dp2[i][j + 1];
      dp2[i][j] %= mod;
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    dp2[m][i] += dp2[m][i + 1];
    dp2[m][i] %= mod;
  }
  long long ans = 0;
  long long ps1 = 0, ps2 = 0;
  for (int i = 1; i <= n; i++) {
    ans += (dp1[m][i] * dp2[m][i]);
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}
