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 p = 1000000007;
long long fact[100000];
long long dp[100000];
long long factdp[100000];
long long ar[200000];
int main() {
  fact[0] = 1;
  fact[1] = 1;
  dp[0] = 1;
  dp[1] = 1;
  factdp[0] = 1;
  factdp[1] = 1;
  for (int i = 2; i < 10000; i++) {
    dp[i] = dp[p % i] * (p - (p / i)) % p;
    fact[i] = i * fact[i - 1] % p;
    factdp[i] = factdp[i - 1] * (dp[i]) % p;
  }
  long long n, m;
  cin >> n >> m;
  long long ans =
      (fact[2 * m + n - 1] * factdp[2 * m] % p * factdp[n - 1] % p + p) % p;
  cout << ans;
}
