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;
void print(vector<long long> A) {
  for (auto x : A) {
    cout << x << " ";
  }
  cout << "\n";
}
void solve(int n, int m) {
  m *= 2;
  vector<vector<long long>> dp(m + 1, vector<long long>(n + 2, 0));
  for (int i = 1; i <= n; i++) {
    dp[1][i] = 1;
  }
  for (int i = 2; i <= m; i++) {
    for (int j = n; j >= 1; j--) {
      dp[i][j] = (dp[i][j + 1] + dp[i - 1][j]) % (long long)(1e9 + 7);
    }
  }
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    res = (res + dp[m][i]) % (long long)(1e9 + 7);
  }
  cout << res << "\n";
}
int main() {
  int n, m;
  cin >> n >> m;
  solve(n, m);
  return 0;
}
