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 f[2005][2005];
long long s[2005][2005], s1[2005][2005];
long long g[2005][2005];
int n, m;
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 2; i <= m; i++) {
    f[1][i] = 1;
    s[1][i] = s[1][i - 1] + 1;
    s1[1][i] = (s1[1][i - 1] + (1 - i) * f[1][i] % 1000000007 + 1000000007) %
               1000000007;
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 2; j <= m; j++) {
      f[i][j] = (j * s[i - 1][j] + s1[i - 1][j]) % 1000000007;
      s[i][j] = (s[i][j - 1] + f[i][j]) % 1000000007;
      s1[i][j] = (s1[i][j - 1] + (1 - j) * f[i][j] % 1000000007 + 1000000007) %
                 1000000007;
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) g[i][j] = (g[i - 1][j] + f[i][j]) % 1000000007;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++)
      ans = (ans + (m - j + 1) * g[i][j] % 1000000007 * f[n - i + 1][j] %
                       1000000007) %
            1000000007;
  printf("%lld", ans);
  return 0;
}
