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];
void prep() {
  for (int e = 2; e < 2010; e++) {
    dp[e][1] = 1;
  }
  for (int t = 2; t < 2010; t++) {
    dp[2][t] = 1;
    long long add = 1;
    for (int e = 3; e < 2010; e++) {
      add = (dp[e][t - 1] + add) % 1000000007;
      dp[e][t] = (dp[e - 1][t] + add) % 1000000007;
    }
  }
  for (int l = 2; l < 2010; l++) sum[l][1] = dp[l][1];
  for (int r = 2; r < 2010; r++) {
    for (int l = 2; l < 2010; l++)
      sum[r][l] = (sum[r][l - 1] + dp[r][l]) % 1000000007;
  }
}
int n, m;
int main() {
  prep();
  while (cin >> n >> m) {
    long long ans = 0ll;
    for (int r = 1; r <= n; r++) {
      for (int l = 2; l <= m; l++) {
        ans = (ans + ((sum[l][r] * (sum[l][n + 1 - r] - sum[l][n - r])) %
                      1000000007) *
                         (m + 1 - l)) %
              1000000007;
      }
    }
    cout << ans << endl;
  }
}
