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;
int N, M;
long long dp[15][(int)(1e3 + 5)];
long long ans;
void printDP() {
  for (int i = 1; i <= M; i++) {
    for (int j = 1; j <= N; j++) {
      cout << setw(15) << dp[i][j];
    }
    cout << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M;
  for (int i = 1; i <= M; i++) {
    for (int j = 0; j <= N; j++) {
      if (i == 1 || j == 1) {
        dp[i][j] = 1;
      } else {
        dp[i][j] =
            dp[i - 1][j] % (int)(1e9 + 7) + dp[i][j - 1] % (int)(1e9 + 7);
      }
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N + 1 - i; j++) {
      ans += dp[M][i] * dp[M][j];
      ans %= (int)(1e9 + 7);
    }
  }
  cout << ans << "\n";
  return 0;
}
