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 MOD = 1000000007;
const int N = 2010;
int n, m, f[N][N], ans;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1, t, s; i <= n; i++) {
    t = 0, s = 1;
    for (int j = 2; j <= m; j++)
      t = (t + f[i - 1][j]) % MOD, f[i][j] = s = (s + t) % MOD;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++)
      ans = (ans + 1ll * (m - j + 1) * (f[i][j] - f[i - 1][j]) % MOD *
                       f[n - i + 1][j]) %
            MOD;
  printf("%d\n", (ans + MOD) % MOD);
  return 0;
}
