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 N = 2005, MD = 1000000007;
int f[N][N], g[N][N];
void add(int &x, int y) {
  x += y;
  if (x >= MD) x -= MD;
}
int main() {
  int i, j, n, m;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; ++i) {
    int sum = 0;
    for (j = 2; j <= m; ++j) {
      add(f[i][j], g[i - 1][j] + 1);
      add(sum, f[i][j]);
      add(g[i][j], g[i][j - 1] + sum);
    }
  }
  long long ans = 0ll;
  for (i = 1; i <= n; ++i) {
    for (j = 2; j <= m; ++j)
      ans += (long long)f[i][j] * (g[n - i][j] - f[n - i][j] + 1) % MD *
             (m - j + 1) % MD;
    ans = ans % MD;
  }
  printf("%I64d", ans);
  return 0;
}
