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 f[2005][2005] = {0};
int g[2005][2005] = {0};
int h[2005][2005] = {0};
int k[2005][2005] = {0};
int n, m;
int main() {
  int ans = 0;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    for (int l = 2; l <= m; l++) {
      k[i][l] = f[i][l] =
          (1ll * (l + 1) * f[i - 1][l] - g[i - 1][l] + 1) % 1000000007;
      h[i][l] =
          (1ll * (l + 1) * f[i - 1][l - 1] - g[i - 1][l - 1] + 1) % 1000000007;
    }
    for (int l = 2; l <= m; l++) {
      g[i][l] = (g[i][l - 1] + 1ll * l * f[i][l]) % 1000000007;
      f[i][l] = (f[i][l - 1] + f[i][l]) % 1000000007;
    }
  }
  for (int i = 1; i <= n; i++)
    for (int l = 2; l <= m; l++) {
      ans = (ans + 1ll * (m - l + 1) *
                       (1ll * k[i][l] * h[n - i + 1][l] % 1000000007)) %
            1000000007;
    }
  printf("%d\n", (ans + 1000000007) % 1000000007);
  return 0;
}
