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 double PI = 3.14159265358979323846;
const long long inf = 1e9, mod = 1e9 + 7, MAX = 1e3 + 1;
long long n, m, dp[21][MAX];
long long sol(int idx, int prev) {
  if (idx == m) return 1;
  long long &ret = dp[idx][prev];
  if (~ret) return ret;
  ret = 0;
  for (int i = prev; i < n + 1; i++) ret += sol(idx + 1, i) % mod, ret %= mod;
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  m *= 2;
  memset(dp, -1, sizeof(dp));
  cout << sol(0, 1) % mod;
}
