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 = 5e5 + 5;
long long fact[N];
long long mod = 1e9 + 7;
long long fastP(long long b, long long e) {
  if (!e) return 1;
  if (e & 1) return b * 1ll * fastP(b * 1ll * b % mod, e >> 1) % mod;
  return fastP(b * 1ll * b % mod, e >> 1) % mod;
}
long long mod_inverse(long long x) { return fastP(x, mod - 2); }
long long nCr(long long n, long long r) {
  if (r > n) return 0;
  return (1LL * fact[n] % mod * mod_inverse(fact[n - r]) % mod *
          mod_inverse(fact[r]) % mod) %
         mod;
}
void fac() {
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = (1LL * i * fact[i - 1]) % mod;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fac();
  long long n, m;
  cin >> n >> m;
  cout << nCr(n + 2 * m - 1, 2 * m) << "\n";
}
