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 MOD = 1000 * 1000 * 1000 + 7;
long long N, M;
long long nbCombi[1010][11];
void construire() {
  for (long long i = 1; i <= N; i++) {
    nbCombi[i][0] = 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++) {
        nbCombi[j][i] = (nbCombi[j][i] + nbCombi[k][i - 1]) % MOD;
      }
    }
  }
}
int32_t main() {
  cin >> N >> M;
  construire();
  long long reponse = 0;
  for (long long gauche = 1; gauche <= N; gauche++) {
    for (long long droite = gauche; droite <= N; droite++) {
      reponse =
          (reponse + nbCombi[gauche][M - 1] * nbCombi[N - droite + 1][M - 1]) %
          MOD;
    }
  }
  cout << reponse << endl;
}
