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 long long INF = 1000000000;
struct edge {
  long long to, cost;
  edge(int a, int b) : to(a), cost(b) {}
};
long long n, m;
long long mod = 1000000007;
long long dp[15][1005][1005];
long long dfs(long long pos, long long sta, long long stb) {
  if (sta > stb) return 0;
  if (pos >= m) return 1;
  if (dp[pos][sta][stb]) return dp[pos][sta][stb];
  long long re = dfs(pos + 1, sta, stb);
  re += dfs(pos, sta, stb - 1) + dfs(pos, sta + 1, stb) -
        dfs(pos, sta + 1, stb - 1) + mod;
  re %= mod;
  dp[pos][sta][stb] = re;
  return re;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  cin >> n >> m;
  cout << dfs(0, 1, n) << endl;
}
