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 inf = 1e9;
const long long inf_ll = 1e18;
const int N = 2005;
const long long mod = 1e9 + 7;
long long dp[N][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    long long x = 0, y = mod - 1;
    for (int j = 1; j <= m; j++) {
      x = (x + dp[i - 1][j]) % mod;
      y = (y + dp[i - 1][j] * j) % mod;
      if (j != 1) dp[i][j] = ((j + 1) * x + mod - y) % mod;
    }
  }
  long long out = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++)
      out = (out + (dp[i][j] + mod - dp[i - 1][j]) * dp[n - i + 1][j] % mod *
                       (m - j + 1)) %
            mod;
  cout << out << "\n";
}
