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 mod = 1000000007;
int n, m;
int f[2333][2333];
int f1[2333][2333];
int sum[2333][2333];
int g[2333];
int main() {
  cin >> n >> m;
  for (int i = 2; i <= m; i++) f[1][i] = 1;
  for (int h = 2; h <= n; h++) {
    int s = 0;
    for (int l = 2; l <= m; l++) {
      s = (s + f[h - 1][l]) % mod;
      f[h][l] = (f[h][l - 1] + s) % mod;
    }
  }
  f[0][0] = 1;
  for (int l = 0; l <= m; l++)
    for (int h = 1; h <= n; h++) f[h][l] = (f[h][l] + f[h - 1][l]) % mod;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= m; j++) sum[i][j] = f[i][j];
  for (int i = 0; i <= n; i++)
    for (int j = 1; j <= m; j++) sum[i][j] = (sum[i][j] + sum[i][j - 1]) % mod;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= m; j++) f1[i][j] = f[i][j];
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int s = 0;
    int last = 0;
    for (int l = 2; l <= m; l++) {
      s = (s + f1[i - 1][l]) % mod;
      last = (last + s) % mod;
      ans = (ans + (m + 1 - l) * (long long)f[i - 1][0] % mod *
                       f[n - i + 1][l] % mod) %
            mod;
      ans = (ans + (m + 1 - l) * (long long)(last - f[i - 1][l]) % mod *
                       f[n - i + 1][l] % mod) %
            mod;
    }
  }
  ans = (ans + mod) % mod;
  cout << ans << endl;
  return 0;
}
