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;
struct _ {
  ios_base::Init i;
  _() { ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL); }
} _;
const int MOD = 1000000007;
int add(int a, int b) {
  a += b;
  if (a >= MOD) a %= MOD;
  return a;
}
int main() {
  int n, m;
  cin >> n >> m;
  vector<vector<int>> dp(2 * m + 1, vector<int>(n + 1, 1));
  int ans = 1;
  for (int i = 2; i <= 2 * m; i++) {
    for (int j = 2; j <= n; j++) {
      dp[i][j] = add(dp[i - 1][j], dp[i][j - 1]);
      if (i == 2 * m) ans = add(ans, dp[i][j]);
    }
  }
  cout << ans;
  return 0;
}
