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 module = 1e9 + 7;
int n, m;
int cnt[2222][2222], cnth[2222][2222];
int sum[2222][2222];
int main() {
  for (int i = 2; i <= 2000; i++) {
    cnt[1][i] = (1LL * i * (i - 1) / 2) % module;
    sum[1][i] = (sum[1][i - 1] + cnt[1][i]) % module;
  }
  for (int i = 2; i <= 2000; i++)
    for (int j = 2; j <= 2000; j++) {
      cnt[i][j] = (cnt[i][j - 1] + sum[i - 1][j]) % module;
      sum[i][j] = (sum[i][j - 1] + cnt[i][j]) % module;
    }
  for (int i = 1; i <= 2000; i++)
    for (int j = 0; j <= 2000; j++)
      cnth[i][j] = (cnth[i - 1][j] + cnt[i][j]) % module;
  cin >> n >> m;
  int ans = 0;
  for (int i = 0; i < n; i++)
    for (int l = 2; l <= m; l++) {
      int s = 1;
      s = (s + cnth[n - i - 1][l]) % module;
      int r = (2LL * cnth[i][l - 1]) % module - cnth[i][l - 2];
      while (r < 0) r += module;
      s = (s + r) % module;
      s = (s + (1LL * r * cnth[n - i - 1][l]) % module) % module;
      ans = (ans + (1LL * (m - l + 1) * s) % module) % module;
    }
  cout << ans << endl;
  return 0;
}
