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;
long long s1[2100][2100], sum[2100][2100], cnt[2100][2100], cnt1[2100][2100];
int main() {
  int n, m, i, j;
  long long res;
  scanf("%d%d", &n, &m);
  for (i = 2; i <= m; i++) {
    for (j = 1; j <= n; j++) {
      cnt[i][j] = sum[i][j - 1] + 1;
      cnt[i][j] %= 1000000007;
      s1[i][j] = s1[i - 1][j] + cnt[i][j];
      s1[i][j] %= 1000000007;
      sum[i][j] = sum[i - 1][j] + s1[i][j];
      sum[i][j] %= 1000000007;
      cnt1[i][j] = cnt[i][j] - cnt[i][j - 1];
      if (cnt1[i][j] < 0) cnt1[i][j] += 1000000007;
    }
  }
  res = 0;
  for (i = 2; i <= m; i++)
    for (j = 1; j <= n; j++) {
      res += ((cnt[i][j] * cnt1[i][n - j + 1]) % 1000000007) * (m - i + 1);
      res %= 1000000007;
    }
  printf("%I64d\n", res);
  return 0;
}
