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;
const long long maxn = 1000 + 10;
const long long maxm = 10 + 10;
const long long Mod = 1e9 + 7;
long long n, m, dp1[maxm][maxn], dp2[maxm][maxn], sigma[maxn];
int main() {
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) dp1[1][i] = dp2[1][i] = 1;
  for (long long i = 1; i <= m; i++) {
    for (long long j = 1; j <= n; j++) {
      for (long long k = 1; k <= j; k++) {
        dp1[i][j] = (dp1[i][j] + dp1[i - 1][k]) % Mod;
        ;
      }
    }
    for (long long j = 1; j <= n; j++) {
      for (long long k = j; k <= n; k++) {
        dp2[i][j] = (dp2[i][j] + dp2[i - 1][k]) % Mod;
      }
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    for (long long j = i; j <= n; j++) {
      ans = (ans + dp1[m][i] * dp2[m][j] % Mod) % Mod;
    }
  }
  cout << ans << endl;
  return 0;
}
