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 int MOD = 1e9 + 7;
int D1[11][1002], D2[11][1002];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int N, M, ans = 0;
  cin >> N >> M;
  D1[0][1] = D2[0][N] = 1;
  for (int i = 1; i <= M; i++) {
    for (int j = 1; j <= N; j++) D1[i][j] = (D1[i - 1][j] + D1[i][j - 1]) % MOD;
    for (int j = N; j > 0; j--) D2[i][j] = (D2[i - 1][j] + D2[i][j + 1]) % MOD;
  }
  ans = (1LL * D1[M][1] * D2[M][1]) % MOD;
  for (int i = 2; i <= N; i++) {
    D1[M][i] = (D1[M][i] + D1[M][i - 1]) % MOD;
    ans = (ans + 1LL * D1[M][i] * D2[M][i]) % MOD;
  }
  cout << ans << '\n';
  return 0;
}
