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 modulo = 1000000007;
const int tope = 2001;
int n, m;
long long int cuantas[tope][tope];
long long int cuantase[tope][tope];
long long int cbis[tope][tope];
long long int cbise[tope][tope];
int main() {
  cin >> n >> m;
  for (int j = 2; j <= m; j++) {
    cuantas[1][j] = 1;
    cuantase[1][j] = 1;
  }
  for (int i = 2; i <= n; i++) {
    long long int amounte = 0;
    for (int j = 2; j <= m; j++) {
      cuantase[i][j] =
          (cuantase[i][j - 1] + amounte + cuantase[i - 1][j]) % modulo;
      cuantas[i][j] = (cuantase[i][j - 1] + amounte) % modulo;
      amounte = (amounte + cuantase[i - 1][j]) % modulo;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 2; j <= m; j++) {
      cbis[i][j] = (cbis[i - 1][j] + cuantas[i][j]) % modulo;
      cbise[i][j] = (cbise[i - 1][j] + cuantase[i][j]) % modulo;
    }
  }
  long long int total = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++)
      total =
          (total + cbis[i][j] * cbise[n - i + 1][j] % modulo * (m - j + 1)) %
          modulo;
  cout << total << endl;
}
