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 int MOD = 1000000007;
int main() {
  int n, m;
  cin >> n >> m;
  if (m == 1) {
    cout << 0 << endl;
    return 0;
  }
  vector<vector<long long> > F(n + 1, vector<long long>(m + 1, 0));
  for (int w = 0; w < (int)(m + 1); ++w) F[0][w] = 1;
  for (int h = (int)(1); h <= (int)(n); ++h) {
    F[h][2] = F[h - 1][2] + 1;
    long long S = F[h - 1][2];
    for (int w = (int)(3); w <= (int)(m); ++w) {
      F[h][w] = (F[h][w - 1] + F[h - 1][w] + S) % MOD;
      S += F[h - 1][w];
    }
  }
  long long res = 0;
  for (int i = 0; i < (int)(n); ++i)
    for (int j = (int)(2); j <= (int)(m); ++j) {
      if (i == 0) {
        res += (long long)(m - j + 1) * F[n - i - 1][j];
      } else {
        res += (long long)(m - j + 1) * (F[i][j] - F[i - 1][j] + MOD) % MOD *
               F[n - i - 1][j];
      }
      res %= MOD;
    }
  cout << res << endl;
}
