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;
vector<long long> adj[1000000];
long long fact[1000000];
void factorial() {
  fact[0] = 1;
  for (long long i = 1; i <= 100000; i++) {
    fact[i] = (fact[i - 1] % 1000000007 * i % 1000000007) % 1000000007;
  }
}
long long power(long long a, long long x) {
  a %= 1000000007;
  long long res = 1;
  while (x > 0) {
    if (x & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    x >>= 1;
  }
  return res;
}
long long nCr(long long n, long long r) {
  long long a = fact[n];
  long long b = power(fact[n - r], 1000000007 - 2);
  long long c = power(fact[r], 1000000007 - 2);
  return (a * (b % 1000000007 * c % 1000000007) % 1000000007) % 1000000007;
}
void solve() {
  factorial();
  long long n, m, i, j, ans = 0;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    for (long long j = n; j >= i; j--) {
      ans += (nCr(m + i - 2, i - 1) % 1000000007 * nCr(m + n - j - 1, n - j) %
              1000000007) %
             1000000007;
    }
  }
  cout << ans % 1000000007;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test = 0, t;
  t = 1;
  while (test++ < t) {
    solve();
  }
}
