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, ans;
int f[2005][2005], g[2005][2005], up[2005][2005], down[2005][2005];
int sf[2005][2005], sg[2005][2005], sup[2005][2005];
int main() {
  cin >> n >> m;
  for (int i = 2; i <= m; i++) {
    f[1][i] = 1;
    g[1][i] = i;
    sf[1][i] = sf[1][i - 1] + f[1][i];
    sg[1][i] = sg[1][i - 1] + g[1][i];
  }
  for (int i = 2; i <= n; i++) {
    f[i][2] = sf[i][2] = 1;
    g[i][2] = sg[i][2] = 2;
    for (int j = 3; j <= m; j++) {
      f[i][j] = ((long long)(j + 1) * sf[i - 1][j] % 1000000007 - sg[i - 1][j] +
                 1000000007) %
                1000000007;
      g[i][j] = (long long)j * f[i][j] % 1000000007;
    }
    for (int j = 3; j <= m; j++) {
      sf[i][j] = (sf[i][j - 1] + f[i][j]) % 1000000007;
      sg[i][j] = (sg[i][j - 1] + g[i][j]) % 1000000007;
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++) {
      up[i][j] = (up[i - 1][j] + f[i][j]) % 1000000007;
      sup[i][j] = (sup[i - 1][j] + sf[i][j]) % 1000000007;
      down[i][j] = (down[i - 1][j] + sg[i][j]) % 1000000007;
    }
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++) {
      long long t1 = up[i][j];
      long long t2 = ((long long)(j + 1) * sup[n - i][j - 1] % 1000000007 -
                      down[n - i][j - 1] + 1000000007) %
                         1000000007 +
                     1;
      ans = (ans + (long long)(m - j + 1) * t1 % 1000000007 * t2 % 1000000007) %
            1000000007;
    }
  printf("%d\n", ans);
}
