Greg has a pad. The pad's screen is an n × m rectangle, each cell can be either black or white. We'll consider the pad rows to be numbered with integers from 1 to n from top to bottom. Similarly, the pad's columns are numbered with integers from 1 to m from left to right.

Greg thinks that the pad's screen displays a cave if the following conditions hold:

  * There is a segment [l, r] (1 ≤ l ≤ r ≤ n), such that each of the rows l, l + 1, ..., r has exactly two black cells and all other rows have only white cells. 
  * There is a row number t (l ≤ t ≤ r), such that for all pairs of rows with numbers i and j (l ≤ i ≤ j ≤ t) the set of columns between the black cells in row i (with the columns where is these black cells) is the subset of the set of columns between the black cells in row j (with the columns where is these black cells). Similarly, for all pairs of rows with numbers i and j (t ≤ i ≤ j ≤ r) the set of columns between the black cells in row j (with the columns where is these black cells) is the subset of the set of columns between the black cells in row i (with the columns where is these black cells). 



Greg wondered, how many ways there are to paint a cave on his pad. Two ways can be considered distinct if there is a cell that has distinct colors on the two pictures.

Help Greg.

Input

The first line contains two integers n, m — the pad's screen size (1 ≤ n, m ≤ 2000).

Output

In the single line print the remainder after dividing the answer to the problem by 1000000007 (109 + 7).

Examples

Input

1 1


Output

0


Input

4 4


Output

485


Input

3 5


Output

451
#include <bits/stdc++.h>
using namespace std;
int n, m;
long long res, c[2005][2005];
int main() {
  cin >> n >> m;
  long long p, q, s;
  for (int i = 1; i <= n; i++) {
    p = 1, q = 0;
    for (int j = 2; j <= m; j++) {
      q = (q + c[i - 1][j]) % 1000000007;
      p = (p + q) % 1000000007;
      c[i][j] = p;
    }
  }
  res = 0;
  for (int i = 1; i < n; i++) {
    p = 1;
    q = 0;
    for (int j = 2; j <= m; j++) {
      s = m - j + 1;
      p = (p + q) % 1000000007;
      q = (q + c[n - i][j]) % 1000000007;
      res = (res + s * (c[i][j] * p % 1000000007)) % 1000000007;
      p = (p + c[n - i][j]) % 1000000007;
    }
  }
  for (int j = 2; j <= m; j++) {
    s = m - j + 1;
    res = (res + s * c[n][j]) % 1000000007;
  }
  cout << res;
}
