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 maxn = 2e5;
const long long mod = 1e9 + 7;
long long n, m;
long long ans;
long long fact[maxn];
void in();
void solve();
void out();
void pp();
long long pwr(long long, long long);
long long C(long long, long long);
int main() {
  in();
  solve();
  out();
}
void in() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  cin >> n >> m;
}
void solve() {
  m = m * 2;
  pp();
  ans = C(m + n - 1, n - 1);
}
void pp() {
  fact[0] = 1;
  for (long long i = 1; i <= n + m; i++) fact[i] = i * fact[i - 1] % mod;
}
void out() { cout << ans; }
long long pwr(long long a, long long b) {
  if (b < 0) return pwr(a, b + mod - 1);
  if (b == 0) return 1;
  long long c = pwr(a, b / 2);
  if (b % 2) return c * c % mod * a % mod;
  return c * c % mod;
}
long long C(long long a, long long b) {
  return fact[a] * pwr(fact[b], -1) % mod * pwr(fact[a - b], -1) % mod;
}
