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;
long long dp[1010][15];
const int mod = 1e9 + 7;
long long dfs(int m, int n) {
  if (dp[n][m]) return dp[n][m];
  if (m == 1) {
    dp[n][m] = n;
    return n;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += dfs(m - 1, i);
    ans %= mod;
  }
  dp[n][m] = ans;
  return ans;
}
int main() {
  int n, m;
  while (cin >> n >> m) {
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      ans =
          (ans + ((dfs(m, i) - dfs(m, i - 1)) * dfs(m, n - i + 1)) % mod) % mod;
      while (ans < 0) ans += mod;
    }
    cout << ans << endl;
  }
  return 0;
}
