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 md = 1000000007;
int n, m;
int f[2010][2010];
int C[6010][4010];
int main() {
  C[0][0] = 1;
  for (int i = 1; i <= 6005; i++) {
    C[i][0] = 1;
    int lim = min(i, 4005);
    for (int j = 1; j <= lim; j++) {
      C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
      if (C[i][j] >= md) C[i][j] -= md;
    }
  }
  scanf("%d %d", &n, &m);
  for (int len = 2; len <= m; len++) f[len][1] = 1;
  for (int h = 2; h <= n; h++) {
    for (int len = 2; len <= m; len++) {
      f[len][h] = C[len + 2 * h - 4][2 * h - 2];
      f[len][h] += f[len][h - 1];
      if (f[len][h] >= md) f[len][h] -= md;
    }
  }
  int ans = 0;
  for (int len = 2; len <= m; len++)
    for (int i = 0; i < (int)(n); i++) {
      long long z =
          (f[len][i + 1] * (long long)(f[len][n - i] - f[len][n - i - 1])) % md;
      z = (z * (m - len + 1)) % md;
      ans = (ans + z) % md;
    }
  printf("%d\n", ans);
  return 0;
}
