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 MAXN = 2000 + 10;
const int mod = 1000000007;
int n, m;
int dp[MAXN][MAXN], func[MAXN][MAXN], C[MAXN][MAXN], f0[MAXN][MAXN],
    f1[MAXN][MAXN], sum[MAXN][MAXN];
int main() {
  for (int i = 0; i < MAXN; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
  }
  cin >> n >> m;
  for (int i = 2; i <= m; i++) {
    dp[1][i] = 1;
    func[1][i] = (func[1][i - 1] + (1 - i)) % mod;
    sum[1][i] = sum[1][i - 1] + 1;
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 2; j <= m; j++) {
      dp[i][j] = ((long long)sum[i - 1][j] * j + func[i - 1][j]) % mod;
      sum[i][j] = (sum[i][j - 1] + dp[i][j]) % mod;
      func[i][j] = (func[i][j - 1] + (long long)(1 - j) * dp[i][j]) % mod;
    }
  }
  for (int j = 2; j <= m; j++)
    for (int i = 1; i <= n; i++) {
      f0[j][i] =
          ((long long)f0[j][i - 1] + (long long)dp[i][j] - dp[i - 1][j]) % mod;
      f1[j][i] = ((long long)f1[j][i - 1] + dp[i][j]) % mod;
    }
  int ret = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++)
      ret = (ret + (long long)(m - j + 1) * (long long)f0[j][i] % mod *
                       f1[j][n - i + 1] % mod) %
            mod;
  ret %= mod;
  ret += mod;
  ret %= mod;
  cout << ret << endl;
  return 0;
}
