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 gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long modexp(long long x, long long n) {
  if (n == 0) {
    return 1;
  } else if (n % 2 == 0) {
    return modexp((x * x) % 1000000007, n / 2);
  } else {
    return (x * modexp((x * x) % 1000000007, (n - 1) / 2)) % 1000000007;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long fact[2003];
  fact[0] = 1;
  for (int i = 1; i <= 2000; i++) {
    fact[i] = i * fact[i - 1];
    fact[i] %= 1000000007;
  }
  long long n, m;
  cin >> n >> m;
  long long res = fact[2 * m + n - 1];
  long long res1 = modexp(fact[2 * m], 1000000007 - 2);
  long long res2 = modexp(fact[n - 1], 1000000007 - 2);
  res *= res1;
  res %= 1000000007;
  res *= res2;
  res %= 1000000007;
  cout << res;
}
