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