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 int jc[12345], jcn[12345];
long long int n, m;
long long int i, j, t, w, h, ans, tmp, pre;
int main() {
  cin >> n >> m;
  for (i = jc[0] = 1; i <= 10000; i++) jc[i] = (jc[i - 1] * i) % 1000000007;
  for (jcn[10000] = 1, j = 1000000007 - 2, i = jc[10000]; j; j >>= 1) {
    if (j & 1) (jcn[10000] *= i) %= 1000000007;
    (i *= i) %= 1000000007;
  }
  for (i = 10000; i; i--) jcn[i - 1] = (jcn[i] * i) % 1000000007;
  ans = 0;
  for (w = 1; w <= m; w++) {
    tmp = pre = 0;
    for (t = 1; t <= n; t++)
      (tmp +=
       ((((jc[t + t + w - 4] * jcn[w - 2]) % 1000000007) * jcn[t + t - 2]) %
        1000000007) *
       ((((jc[n - t + 1 + n - t + 1 + w - 4] * jcn[w - 2]) % 1000000007) *
         jcn[n - t + 1 + n - t + 1 - 2]) %
        1000000007)) %= 1000000007;
    for (t = n - 1, i = 1; t; t--, i++)
      (pre +=
       ((((jc[i + i + w - 4] * jcn[w - 2]) % 1000000007) * jcn[i + i - 2]) %
        1000000007)) %= 1000000007,
          (tmp +=
           ((((jc[t + t + w - 4] * jcn[w - 2]) % 1000000007) * jcn[t + t - 2]) %
            1000000007) *
           pre) %= 1000000007;
    (ans += (m - w + 1) * tmp) %= 1000000007;
  }
  cout << ans;
  return 0;
}
