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 INF = 1E9 + 7;
template <class C>
void mini(C& a4, C b4) {
  a4 = min(a4, b4);
}
template <class C>
void maxi(C& a4, C b4) {
  a4 = max(a4, b4);
}
const long long mod = 1E9 + 7;
long long t[2005][2005];
long long tm[2005][2005];
int main() {
  ios_base::sync_with_stdio(false);
  int h, w;
  cin >> h >> w;
  for (int i = (2); i <= (w); ++i) t[1][i] = tm[1][i] = 1;
  for (int i = (2); i <= (h); ++i) {
    long long s = 0;
    for (int j = (2); j <= (w); ++j) {
      s += t[i - 1][j];
      s %= mod;
      t[i][j] = (t[i][j - 1] + s) % mod;
      tm[i][j] = (tm[i - 1][j] + t[i][j]) % mod;
    }
  }
  long long S = 0;
  for (int x = (1); x <= (h); ++x) {
    for (int szer = (2); szer <= (w); ++szer) {
      long long ile = tm[x][szer] * tm[h - x + 1][szer];
      if (x != 1) ile -= tm[x - 1][szer] * tm[h - x + 1][szer];
      ile %= mod;
      ile *= w - szer + 1;
      ile %= mod;
      S += ile;
      S %= mod;
    }
  }
  cout << S << endl;
  return 0;
}
