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 INF = 0x3f3f3f3f;
const long long int LINF = 0x3f3f3f3f3f3f3f3fll;
mt19937 rng((int)chrono::steady_clock::now().time_since_epoch().count());
const int MAX = (int)1e5 + 10;
const long long int mod = 1e9 + 7;
vector<long long int> fact(10000);
vector<long long int> fact_inv(10000);
long long int binpow(long long int a, long long int b) {
  a %= mod;
  long long int ret = 1;
  while (b > 0) {
    if (b & 1) {
      ret = (ret * a) % mod;
    }
    a = (a * a) % mod;
    b = b >> 1;
  }
  return ret % mod;
}
int main() {
  ios::sync_with_stdio(false);
  ;
  long long int n, m;
  cin >> n >> m;
  fact[0] = 1;
  fact[1] = 1;
  for (long long int i = 2; i < 10000; i++) {
    fact[i] = (fact[i - 1] * i) % mod;
  }
  for (long long int i = 0; i < 10000; i++) {
    fact_inv[i] = binpow(fact[i], mod - 2) % mod;
  }
  m *= 2;
  n += m - 1;
  long long int resp = fact[n];
  resp = (resp * fact_inv[m]) % mod;
  resp = (resp * fact_inv[n - m]) % mod;
  cout << resp << '\n';
  return 0;
}
