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 mod = 1e9 + 7;
long long comb[2000][2000];
long long combi(long long n, long long r) {
  if (n < r || n < 0) return 0;
  if (comb[n][r] != 0) return comb[n][r];
  if (r == 0 || r == n) return comb[n][r] = 1;
  return comb[n][r] = (combi(n - 1, r - 1) + combi(n - 1, r)) % mod;
}
int main() {
  long long n, m;
  cin >> n >> m;
  for (int i = 0; i < 2000; i++) {
    for (int j = 0; j < 2000; j++) {
      comb[i][j] = 0;
    }
  }
  long long cnt = 0;
  for (long long i = 1; i <= n; i++) {
    long long a_cnt = combi(i + m - 1, m) - combi(i + m - 2, m);
    while (a_cnt < 0) a_cnt += mod;
    cnt += a_cnt % mod * combi(n - i + m, m) % mod;
    cnt %= mod;
  }
  cout << cnt << endl;
  return 0;
}
