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>
const int kMaxSize = 2000 + 10;
const int kDiv = 1000000000 + 7;
int n, m;
long long f[kMaxSize][kMaxSize], s[kMaxSize][kMaxSize];
void Initialize() {
  int i, j, k;
  for (i = 2; i <= m; ++i) {
    f[1][i] = 1;
    s[1][i] = s[1][i - 1] + f[1][i];
  }
  for (i = 2; i <= n; ++i) {
    for (j = 2; j <= m; ++j) {
      f[i][j] = (f[i][j - 1] + s[i - 1][j]) % kDiv;
      s[i][j] = (s[i][j - 1] + f[i][j]) % kDiv;
    }
  }
  memset(s, 0, sizeof(s));
  for (i = 1; i <= n; ++i)
    for (j = 2; j <= m; ++j) s[i][j] = (s[i - 1][j] + f[i][j]) % kDiv;
}
int main() {
  scanf("%d %d", &n, &m);
  Initialize();
  long long result = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 2; j <= m; ++j)
      result =
          (result + s[i][j] * (s[n + 1 - i][j] - s[n - i][j]) * (m + 1 - j)) %
          kDiv;
  printf("%I64d\n", result);
  return 0;
}
