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 long long Mod = 1000000007;
int n, m;
long long dp[2020][2020], same_with_last[2020][2020], sum[2020], ans[2020];
long long presum[2020][2020];
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) dp[i][0] = 1;
  for (int i = 0; i <= n; i++) {
    ans[i] = 1;
    for (int j = 0; j <= m; j++) {
      sum[i] = (sum[i] + dp[i][j]) % Mod;
      ans[i] = (ans[i] + sum[i] - dp[i][0] + Mod) % Mod;
      if (j >= 2) {
        dp[i + 1][j] = ans[i], same_with_last[i + 1][j] = dp[i][j];
      }
    }
  }
  for (int val = 2; val <= m; val++) {
    for (int i = 1; i <= n; i++) {
      presum[val][i] =
          (presum[val][i - 1] + dp[i][val] - same_with_last[i][val] + Mod) %
          Mod;
    }
  }
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    for (int val = 2; val <= m; val++) {
      long long now =
          (dp[i][val] - same_with_last[i][val] + Mod) * (m - val + 1) % Mod;
      res = (res + now * presum[val][n - i + 1]) % Mod;
    }
  }
  cout << res << endl;
  return 0;
}
