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 P = 7 + 1e9;
long long ar[2005];
long long fn(long long b, long long p, long long m) {
  if (p == 0) return 1;
  if (p == 1) return b;
  long long x = fn(b, p / 2, m) % P;
  if (p % 2 == 0) return x * x % P;
  x = (fn(b, p - 1, m) % P * b) % P;
  return x;
}
long long ncr(long long n, long long r) {
  long long q = (ar[r] * ar[n - r]) % P;
  q = fn(q, P - 2, P);
  return ar[n] * q % P;
}
int main() {
  ar[0] = 1;
  for (long long i = 1; i < 2005; i++) ar[i] = (ar[i - 1] * i) % P;
  long long n, m;
  cin >> n >> m;
  long long ans = ncr(2 * m + n - 1, n - 1);
  cout << ans;
  return 0;
}
