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 modu = 1000000007LL;
long long dp_freecnt[2050][2050][2];
long long Get(int space, int freedom, int left_tied = 0) {
  if (space < 0) return 0LL;
  long long& alias = dp_freecnt[space][freedom][left_tied];
  if (alias != -1LL) return alias;
  alias = 1LL;
  if (freedom) alias += Get(space, freedom - 1, 0);
  alias %= modu;
  if (!left_tied) alias += Get(space - 1, freedom, left_tied);
  alias %= modu;
  alias += Get(space - 1, freedom, 1);
  alias %= modu;
  return alias;
}
int main() {
  int row, col;
  cin >> row >> col;
  for (int(i) = 0; (i) < (2050); ++(i))
    for (int(j) = 0; (j) < (2050); ++(j))
      for (int(k) = 0; (k) < (2); ++(k)) dp_freecnt[i][j][k] = -1LL;
  long long res = 0LL;
  for (int(i) = 0; (i) < (row); ++(i)) {
    for (int jar = 1; jar < col; ++jar) {
      long long cnt = (long long)(col - jar);
      long long atas = 1LL;
      if (i) {
        atas += Get(jar - 1, i - 1);
        atas %= modu;
      }
      long long bawah = 1LL;
      if (i != row - 1) {
        int sisa = row - i - 2;
        long long mult = Get(jar - 2, sisa) * 2LL - Get(jar - 3, sisa);
        mult %= modu;
        while (mult < modu) mult += modu;
        bawah += mult;
        bawah %= modu;
      }
      cnt *= atas;
      cnt %= modu;
      cnt *= bawah;
      cnt %= modu;
      res += cnt;
      res %= modu;
    }
  }
  cout << res << endl;
}
