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;
void inline add(int &A, int B) {
  A += B;
  if (A >= 1000000007) A -= 1000000007;
}
void inline sub(int &A, int B) {
  A -= B;
  if (A < 0) A += 1000000007;
}
int inline mul(int A, int B) { return (1ll * A * B) % 1000000007; }
int n, m, dp[2005][2005], sum[2005][2005], f[2005][2005];
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 2; j <= m; j++) {
      dp[i][j] = 1;
      add(dp[i][j], mul(j, sum[i - 1][j]));
      sub(dp[i][j], f[i - 1][j]);
      sum[i][j] = dp[i][j];
      f[i][j] = mul(dp[i][j], j - 1);
    }
    for (int j = 2; j <= m; j++) {
      add(sum[i][j], sum[i][j - 1]);
      add(f[i][j], f[i][j - 1]);
    }
  }
  int res = 0;
  for (int up = 1; up <= n; up++)
    for (int val = 2; val <= m; val++) {
      int down = n - up + 1;
      add(res, mul(m - val + 1, mul(dp[up][val], (1000000007 + dp[down][val] -
                                                  dp[down - 1][val]))));
    }
  cout << res;
}
