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 pd[2010][2010][2];
long long spd[2010][2010][2];
int main() {
  int n, m;
  for (long long r = 1; r >= 0; r--)
    for (long long hm = 1; hm <= 2000; hm++) {
      long long acc = 0;
      for (long long w = 2; w <= 2000; w++) {
        if (r)
          acc = (acc + spd[w][hm - 1][1] + pd[w][hm - 1][1]) % 1000000007LL;
        else
          acc = (acc + spd[w][hm - 1][1]) % 1000000007LL;
        pd[w][hm][r] = (1 + acc) % 1000000007LL;
        spd[w + 1][hm][r] = (spd[w][hm][r] + pd[w][hm][r]) % 1000000007LL;
      }
    }
  while (scanf("%d%d", &n, &m) == 2) {
    long long res = 0;
    for (long long w = 2; w <= m; w++)
      for (long long t = 1; t <= n; t++)
        res = (res + (((((pd[w][t][1] - pd[w][t - 1][1] + 1000000007LL) %
                         1000000007LL) *
                        pd[w][n - t + 1][1]) %
                       1000000007LL) *
                      (m - w + 1)) %
                         1000000007LL) %
              1000000007LL;
    printf("%d\n", (int)res);
  }
  return 0;
}
