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 N = 2e3 + 10;
const int BASE = 1e9 + 7;
int n, m;
int dp[N][N];
void Add(int &a, int b) {
  a = (a + b) % BASE;
  if (a >= BASE) a -= BASE;
  if (a < 0) a += BASE;
}
int Mul(int a, int b) { return 1LL * a * b % BASE; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  if (m <= 1) {
    cout << 0;
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    int p = 0, q = 0;
    for (int j = 2; j <= m; ++j) {
      Add(p, dp[i - 1][j]);
      Add(q, Mul(dp[i - 1][j], j));
      dp[i][j] = 1 + Mul(p, j + 1) - q;
      Add(dp[i][j], 0);
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 2; j <= m; ++j) {
      int w = Mul(dp[i][j], dp[n - i + 1][j]);
      w = Mul(w, m - j + 1);
      Add(ans, w);
      if (i < n) {
        w = Mul(dp[i][j], dp[n - i][j]);
        w = Mul(w, m - j + 1);
        Add(ans, -w);
      }
    }
  cout << ans << '\n';
  return 0;
}
