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 n, m, i, j, now1, now2, now, Ans;
long long f[2005][2005], f2[2005][2005];
int main() {
  scanf("%I64d %I64d", &n, &m);
  f[0][0] = 1;
  for (i = 1; i <= n; i++) {
    f[i][0] = 1;
    now1 = 0;
    now2 = 1;
    for (j = 2; j <= m; j++) {
      now1 += f[i - 1][j], now1 %= 1000000007;
      now2 += now1, now2 %= 1000000007;
      f[i][j] = now2;
      f2[i][j] = (f[i][j] - f[i - 1][j]) % 1000000007;
    }
  }
  Ans = 0;
  for (i = 1; i <= n; i++)
    for (j = 2; j <= m; j++) {
      now = f2[i][j];
      now *= f[n - i + 1][j];
      now %= 1000000007;
      now *= (m - j + 1);
      now %= 1000000007;
      Ans += now;
      Ans %= 1000000007;
    }
  while (Ans < 0) Ans += 1000000007;
  printf("%I64d\n", Ans);
  return 0;
}
