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 MAX = 2000 + 10;
const int Mod = 1000000007;
int n, m;
int C[MAX][MAX];
int f[MAX][MAX], f2[MAX][MAX];
int sum[MAX], sum2[MAX];
int main() {
  int i, j, k;
  C[0][0] = 1;
  for ((i) = (1); (i) != (MAX); ++(i)) {
    C[i][0] = 1;
    for ((j) = (1); (j) <= (i); ++(j))
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % Mod;
  }
  scanf("%d%d", &n, &m);
  for ((i) = (2); (i) <= (m); ++(i)) f[1][i] = 1;
  for ((i) = (2); (i) <= (n); ++(i)) {
    for ((k) = (1); (k) <= (m); ++(k)) {
      sum[k] = (sum[k - 1] + (long long)f[i - 1][k] * (k - 1)) % Mod;
      sum2[k] = (sum2[k - 1] + f[i - 1][k]) % Mod;
    }
    for ((j) = (2); (j) <= (m); ++(j))
      f[i][j] = ((long long)sum2[j] * j - sum[j] + Mod) % Mod;
  }
  for ((i) = (2); (i) <= (n); ++(i)) {
    for (k = m; k >= 1; --k) {
      sum[k] =
          (sum[k + 1] + (long long)f[i - 1][k] * (k + 1) * (m - k + 1)) % Mod;
      sum2[k] = (sum2[k + 1] + (long long)f[i - 1][k] * (m - k + 1)) % Mod;
    }
    for ((j) = (2); (j) <= (m); ++(j))
      f2[i][j] = (sum[j + 1] - (long long)sum2[j + 1] * j % Mod + Mod) % Mod;
    for (k = m; k >= 1; --k) {
      sum[k] = (sum[k + 1] + (long long)f2[i - 1][k] * (k + 1)) % Mod;
      sum2[k] = (sum2[k + 1] + f2[i - 1][k]) % Mod;
    }
    for ((j) = (2); (j) <= (m); ++(j))
      f2[i][j] =
          ((f2[i][j] + sum[j]) % Mod - (long long)sum2[j] * j % Mod + Mod) %
          Mod;
  }
  int ans = 0;
  for ((i) = (1); (i) <= (n); ++(i))
    for ((j) = (1); (j) <= (m); ++(j))
      ans = (ans + (long long)(f2[i][j] + (long long)f[i][j] * (m - j + 1)) %
                       Mod * (n - i + 1)) %
            Mod;
  cout << ans << endl;
  return 0;
}
