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;
int n, m, i, j, k, ans, sum;
int f[2005][2005], f2[2005][2005], g[2005][2005];
int main() {
  scanf("%d%d", &n, &m);
  for (i = 2; i <= m; ++i) f[1][i] = 1, g[1][i] = i - 1;
  for (i = 2; i <= n; ++i) {
    sum = 0;
    for (j = 2; j <= m; ++j) {
      sum += g[i - 1][j];
      if (sum >= 1000000007) sum -= 1000000007;
      if (i == 2)
        f[i][j] = sum - 1;
      else
        f[i][j] = sum;
    }
    for (j = 2; j <= m; ++j) {
      g[i][j] = f[i][j] + g[i][j - 1];
      if (g[i][j] >= 1000000007) g[i][j] -= 1000000007;
    }
  }
  for (i = 2; i <= m; ++i) f2[1][i] = 1, g[1][i] = i - 1;
  for (i = 2; i <= n; ++i) {
    sum = 0;
    for (j = 2; j <= m; ++j) {
      sum += g[i - 1][j];
      if (sum >= 1000000007) sum -= 1000000007;
      f2[i][j] = sum;
    }
    for (j = 2; j <= m; ++j) {
      g[i][j] = f2[i][j] + g[i][j - 1];
      if (g[i][j] >= 1000000007) g[i][j] -= 1000000007;
    }
  }
  for (i = 2; i <= n; ++i)
    for (j = 2; j <= m; ++j) f[i][j] += f[i - 1][j];
  for (i = 2; i <= n; ++i)
    for (j = 2; j <= m; ++j) f2[i][j] += f2[i - 1][j];
  for (i = 1; i <= n; ++i)
    for (j = 2; j <= m; ++j) {
      ans +=
          (m - j + 1ll) * f[i][j] % 1000000007 * f2[n - i + 1][j] % 1000000007;
      if (ans >= 1000000007) ans -= 1000000007;
    }
  printf("%d\n", ans);
}
