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;
long long n, m, ans, mod = 1e9 + 7;
long long fexp(long long a, int x, long long mod) {
  if (x == 0) return 1ll;
  if (x % 2 == 0) {
    long long y = fexp(a, x / 2, mod);
    return (y * y) % mod;
  }
  return (a * fexp(a, x - 1, mod)) % mod;
}
long long divv(long long a, long long b, long long mod) {
  return (a * fexp(b, mod - 2, mod)) % mod;
}
long long f[1000100];
long long fat(long long a, long long mod) {
  if (a <= 1) return 1;
  return f[a] ? f[a] : (f[a] = (a * fat(a - 1, mod)) % mod);
}
long long choose(long long n, long long k, long long mod) {
  return divv(fat(n, mod), (fat(k, mod) * fat(n - k, mod)) % mod, mod) % mod;
}
long long fw(long long x) { return choose(m + x - 1, x - 1, mod); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    ans += choose(m + i - 2, i - 1, mod) * fw(n - i + 1);
    ans %= mod;
  }
  cout << ans << endl;
}
