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 int MOD = 1e9 + 7;
const int MX = 100005;
long long int fact[100005];
long long int inv[100005];
long long int finv[100005];
void prec(long long int mod = MOD) {
  fact[0] = inv[0] = finv[0] = 1;
  fact[1] = inv[1] = finv[1] = 1;
  for (int i = 2; i <= MX; i++) {
    fact[i] = (fact[i - 1] * i) % mod;
    inv[i] = mod - (mod / i) * inv[mod % i] % mod;
    finv[i] = (finv[i - 1] * inv[i]) % mod;
  }
}
int main() {
  long long int n, m;
  cin >> n >> m;
  prec();
  long long int x = (2 * m + n - 1);
  long long int y = 2 * m;
  long long int z = n - 1;
  long long int ans = fact[x] * finv[y] % MOD * finv[z] % MOD;
  cout << ans << endl;
}
