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;
long long a[2010][2010], b[2010][2010], s[2010][2010], sum[2010][2010];
int main() {
  int i, j, k, n, m;
  long long ans;
  while (scanf("%d %d ", &n, &m) == 2) {
    for (i = 1; i <= n; i++) {
      for (j = 2; j <= m; j++) {
        a[i][j] = (1 + sum[i - 1][j]) % 1000000007;
        s[i][j] = (s[i][j - 1] + a[i][j]) % 1000000007;
        sum[i][j] = (sum[i][j - 1] + s[i][j]) % 1000000007;
        b[i][j] = (a[i][j] - a[i - 1][j] + 1000000007) % 1000000007;
      }
    }
    ans = 0;
    for (i = 1; i <= n; i++)
      for (j = 2; j <= m; j++)
        ans = (ans + a[i][j] * b[n - i + 1][j] % 1000000007 * (m - j + 1) %
                         1000000007) %
              1000000007;
    printf("%I64d\n", ans);
  }
  return 0;
}
