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;
struct timer {
  time_t start;
  timer() { start = clock(); }
  ~timer() {
    cerr << 1. * (clock() - start) / CLOCKS_PER_SEC << " secs" << endl;
  }
};
int n, m;
long long dp1[2010][2010];
long long sumJ[2010][2010];
long long sumI1[2010][2010];
long long dp2[2010][2010];
long long sumI2[2010][2010];
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for (int j = 2; j < (m + 1); j++) {
    dp1[1][j] = dp2[1][j] = 1;
    sumJ[1][j] = sumJ[1][j - 1] + dp1[1][j];
  }
  for (int i = 2; i < (n + 1); i++)
    for (int j = 2; j < (m + 1); j++) {
      dp1[i][j] = (dp1[i][j - 1] + sumJ[i - 1][j]) % 1000000007LL;
      sumJ[i][j] = (sumJ[i][j - 1] + dp1[i][j]) % 1000000007LL;
      dp2[i][j] = (dp1[i][j] - dp1[i - 1][j] + 1000000007LL) % 1000000007LL;
    }
  for (int i = 1; i < (n + 1); i++)
    for (int j = 2; j < (m + 1); j++) {
      sumI1[i][j] = (sumI1[i - 1][j] + dp1[i][j]) % 1000000007LL;
      sumI2[i][j] = (sumI2[i - 1][j] + dp2[i][j]) % 1000000007LL;
    }
  long long ans = 0;
  for (int i = 1; i < (n + 1); i++)
    for (int j = 2; j < (m + 1); j++) {
      ans += sumI1[i][j] * sumI2[n + 1 - i][j] % 1000000007LL * (m + 1 - j);
      ans %= 1000000007LL;
    }
  cout << ans;
}
