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 dp[2005][2005];
int n, m;
long long ans = 0;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) dp[1][i] = 1;
  for (int i = 2; i <= n; i++) {
    int sum = 0;
    dp[i][1] = 1;
    for (int r = 2; r <= m; r++)
      sum += dp[i - 1][r], sum %= 1000000007,
          dp[i][r] = (dp[i][r - 1] + sum) % 1000000007;
  }
  for (int i = 1; i <= n; i++)
    for (int r = 2; r <= m; r++)
      ans = (ans + 1000000007 +
             ((dp[i][r] - dp[i - 1][r]) * dp[n - i + 1][r] % 1000000007) *
                 (m - r + 1)) %
            1000000007;
  cout << ans;
}
