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 int N = 2002;
long long fact[N];
long long powerMod(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = ((res % 1000000007) * (a % 1000000007)) % 1000000007;
    a = ((a % 1000000007) * (a % 1000000007)) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long modInverse(long long n, long long m) { return powerMod(n, m - 2); }
long long nCr(long long n, long long r) {
  long long temp = fact[n];
  (temp *=
   (modInverse(fact[r], 1000000007) * modInverse(fact[n - r], 1000000007)) %
   1000000007) %= 1000000007;
  return temp;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed;
  cout.precision(10);
  ;
  long long n, m;
  cin >> n >> m;
  fact[0] = fact[1] = 1;
  for (long long i = 2; i < N; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000007;
  }
  cout << nCr(n + (2 * m) - 1, 2 * m);
  return 0;
}
