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>
#pragma warning(disable : 4996)
long long D[2010][2010], Mod = 1000000007, Res, D2[2010][2010], D3[2010][2010],
                         S;
int n, m;
int main() {
  int i, j;
  scanf("%d%d", &n, &m);
  for (i = 2; i <= m; i++) {
    D2[0][i] = 1;
    D[1][i] = i - 1;
    D3[1][i] = 1;
  }
  for (i = 2; i <= n; i++) {
    S = 0;
    for (j = 2; j <= m; j++) {
      D3[i][j] = S + D[i - 1][j - 1];
      S = (S + D[i - 1][j]) % Mod;
      D[i][j] = (D[i][j - 1] + S) % Mod;
    }
  }
  for (i = 2; i <= n; i++) {
    for (j = 2; j <= m; j++) {
      D3[i][j] = (D3[i - 1][j] + D3[i][j]) % Mod;
    }
  }
  for (i = 1; i <= n; i++) {
    S = 0;
    for (j = 1; j <= m; j++) {
      S += D[i][j];
      D2[i][j] = (D2[i - 1][j] + S + Mod) % Mod;
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      Res = (Res + D3[i][j] * D2[n - i][j] % Mod * (m - j + 1)) % Mod;
    }
  }
  printf("%lld\n", Res);
}
