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 MOD = 1e9 + 7;
const long long N = 2000;
long long fact[N];
void init() {
  fact[0] = 1;
  for (long long i = 1; i < N; i++) fact[i] = fact[i - 1] * i % MOD;
}
long long fastpow(long long n, long long p) {
  long long res = 1;
  while (p) {
    if (p & 1) res = (res * n) % MOD;
    n = n * n % MOD;
    p >>= 1;
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  init();
  long long n, m;
  cin >> n >> m;
  cout << fact[n + 2 * m - 1] * fastpow(fact[2 * m], MOD - 2) % MOD *
              fastpow(fact[n - 1], MOD - 2) % MOD
       << "\n";
}
