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[2010][2010];
long long sum[2010][2010][2];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int j = 2; j <= m; j++) {
    dp[1][j] = 1;
    sum[1][j][0] = j - 1;
    sum[1][j][1] = sum[1][j - 1][1] + j;
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 2; j <= m; j++) {
      dp[i][j] = sum[i - 1][j][0] * (j + 1) - sum[i - 1][j][1];
      dp[i][j] = dp[i][j] % 1000000007;
      if (dp[i][j] < 0) dp[i][j] += 1000000007;
      sum[i][j][0] = (sum[i][j - 1][0] + dp[i][j]) % 1000000007;
      sum[i][j][1] = (sum[i][j - 1][1] + dp[i][j] * j) % 1000000007;
      dp[i][j] = dp[i][j] + dp[i - 1][j];
      if (dp[i][j] >= 1000000007) dp[i][j] -= 1000000007;
    }
  }
  long long tot = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++) {
      tot += dp[i][j] * dp[n - i + 1][j] % 1000000007 * (m - j + 1);
      tot -= dp[i][j] * dp[n - i][j] % 1000000007 * (m - j + 1);
      tot %= 1000000007;
    }
  tot = (tot % 1000000007 + 1000000007) % 1000000007;
  printf("%d\n", tot);
}
