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;
const int N = 2005, Mod = 1000000007;
int f[N][N], g[N][N], sum, Ans, n, m;
inline int Read() {
  int t = 0, f = 1;
  char c = getchar();
  for (; c > 57 || c < 48; c = getchar())
    if (c == '-') f = -1;
  for (; c > 47 && c < 58; c = getchar()) t = (t << 1) + (t << 3) + c - 48;
  return t * f;
}
inline int Add(int &x, int y) { x = (x + y % Mod) % Mod; }
int main() {
  n = Read(), m = Read();
  for (int i = 1; i <= n; i++, sum = 0)
    for (int j = 2; j <= m; j++) {
      Add(f[i][j], g[i - 1][j] + 1);
      Add(sum, f[i][j]), Add(g[i][j], g[i][j - 1] + sum);
    }
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++)
      Add(Ans, 1ll * f[i][j] * (g[n - i][j] - f[n - i][j] + 1) % Mod *
                   (m - j + 1) % Mod);
  printf("%d\n", Ans);
}
