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 int N = 1e3 + 5;
const int mod = 1e9 + 7;
int n, m;
long long dp[15][N];
long long solve(int x, int a) {
  long long &cache = dp[x][a];
  if (cache) return cache;
  if (x == m) return cache = (n - a + 1);
  long long ans = 0;
  for (int i = int(a); i <= int(n); i++) {
    ans = (ans + (i - a + 1) * solve(x + 1, i)) % mod;
  }
  return cache = ans;
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  solve(0, 1);
  long long ans = 0;
  for (int i = int(1); i <= int(n); i++) {
    ans = (ans + dp[1][i]) % mod;
  }
  cout << ans << '\n';
  return 0;
}
