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>
int n, m;
int f[2222][2222], s[2222][2222];
int add(int a, int b) {
  return a + b >= 1000000007 ? a + b - 1000000007 : a + b;
}
int mul(int a, int b) { return 1LL * a * b % 1000000007; }
int main() {
  scanf("%d%d", &n, &m);
  memset(f, 0, sizeof f);
  memset(s, 0, sizeof s);
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++) {
      f[i][j] = (j == 2) ? 1 : 0;
      f[i][j] = add(f[i][j], f[i][j - 1]);
      f[i][j] = add(f[i][j], s[i - 1][j]);
      s[i][j] = add(s[i][j - 1], f[i][j]);
    }
  int res = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++) {
      int t = m - j + 1;
      t = mul(t, f[i][j]);
      t = mul(t, add(f[n - i + 1][j], 1000000007 - f[n - i][j]));
      res = add(res, t);
    }
  printf("%d\n", res);
  return 0;
}
