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 maxn = 2e3 + 7;
long long read() {
  long long res = 0, f = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (isdigit(c)) res = (res << 1) + (res << 3) + c - 48, c = getchar();
  return res * f;
}
long long ans;
int n, m;
long long f[maxn][maxn], g[maxn][maxn];
int main() {
  n = read(), m = read();
  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] = (f[i - 1][j] + 2 * f[i][j - 1] - f[i][j - 2] + 1000000009) %
                1000000009;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++) g[i][j] = (g[i - 1][j] + f[i][j]) % 1000000009;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++)
      (ans += f[i][j] * g[n - i + 1][j] % 1000000009 * (m - j + 1) %
              1000000009) %= 1000000009;
  printf("%lld\n", ans);
  return 0;
}
