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>
const int mod = 1e9 + 7;
long long factorial[30001];
long long fast(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % mod;
    }
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long inv(long long x) { return fast(x, mod - 2); }
long long C(long long a, long long b) {
  long long res = factorial[a];
  res = (res * inv(factorial[b])) % mod;
  res = (res * inv(factorial[a - b])) % mod;
  return res;
}
int main() {
  factorial[0] = 1;
  for (long long i = 1; i <= 30000; i++) {
    factorial[i] = (factorial[i - 1] * i) % mod;
  }
  long long n, m;
  scanf("%lld%lld", &n, &m);
  printf("%lld\n", C(n + 2 * m - 1, 2 * m));
}
