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;
int n, m, i, j;
int f[2222][2222];
int t[2222][2222];
int s[2222][2222];
int *ff, *ss, *tt;
int ans, cur;
int main() {
  scanf("%d%d", &n, &m);
  if (m == 1) {
    printf("0\n");
    return 0;
  }
  ff = f[0];
  tt = t[0];
  for (i = 1; i < m; i++) {
    ff[i] = 1;
    tt[i] = 1;
  }
  for (i = 1; i < n; i++) {
    ss = s[i];
    ff = f[i];
    tt = t[i];
    ss[0] = 0;
    ff[0] = 0;
    for (j = 1; j < m; j++) {
      ss[j] = (ss[j - 1] + f[i - 1][j]) % 1000000007;
      ff[j] = (ff[j - 1] + ss[j]) % 1000000007;
      tt[j] = (t[i - 1][j] + ff[j]) % 1000000007;
    }
  }
  ans = 0;
  for (i = 1; i < m; i++) {
    cur = 0;
    for (j = 0; j < n; j++)
      cur = (cur + (int)(1ll * f[j][i] * t[n - j - 1][i] % 1000000007)) %
            1000000007;
    ans = (ans + (int)((m - i | 0ll) * cur % 1000000007)) % 1000000007;
  }
  printf("%d", ans);
  return 0;
}
