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 int mod = 1e9 + 7;
int Dec[11][1001], Inc[11][1001], m, n;
int cal_inc(int pos, int last) {
  if (pos == m) return 1;
  int &ret = Inc[pos][last];
  if (ret != -1) return ret;
  ret = 0;
  for (int val = last; val <= n; val++)
    ret = (ret + cal_inc(pos + 1, val)) % mod;
  return ret;
}
int cal_dec(int pos, int last) {
  if (pos == 1) return 1;
  int &ret = Dec[pos][last];
  if (ret != -1) return ret;
  ret = 0;
  for (int val = last; val >= 1; val--)
    ret = (ret + cal_dec(pos - 1, val)) % mod;
  return ret;
}
int main() {
  cin >> n >> m;
  memset(Dec, -1, sizeof(Dec));
  memset(Inc, -1, sizeof(Inc));
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      ans = (ans + cal_dec(m, i) * 1LL * cal_inc(1, j) % mod) % mod;
    }
  }
  cout << ans;
  return 0;
}
