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 n, m;
long long mod = 1e9 + 7;
long long f[2100], q[2100];
long long pow(long long a, long long b) {
  if (b == 0)
    return 1ll;
  else if (b % 2 == 1)
    return (pow(a, b - 1) % mod * a % mod) % mod;
  else {
    long long g = pow(a, b / 2);
    return (g % mod * g % mod) % mod;
  }
}
long long c(long long n, long long k) {
  return (q[n] % mod * pow(f[k], mod - 2) % mod) % mod;
}
int main() {
  cin >> n >> m;
  f[0] = 1;
  for (long long i = 1; i <= 2000; i++) {
    f[i] = (f[i - 1] * i) % mod;
  }
  q[n] = n;
  for (long long i = n + 1; i <= 2000; i++) {
    q[i] = (q[i - 1] * i) % mod;
  }
  cout << c(n + 2 * m - 1, 2 * m);
  return 0;
}
