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 = 2010, mod = 1LL * 1000 * 1000 * 1000 + 7;
int dp[N][N], n, m, s[N][N], sum[N][N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 2; i < N; i++) dp[i][0] = 1;
  for (int i = 1; i < N; i++)
    for (int j = i; j >= 2; j--) {
      (sum[i][0] += dp[j][0]) %= mod;
      (s[i][0] += 1LL * dp[j][0] * (i - j + 1) % mod) %= mod;
    }
  for (int j = 1; j < N; j++)
    for (int i = 2; i < N; i++) {
      dp[i][j] = 1;
      (dp[i][j] += s[i][j - 1]) %= mod;
      sum[i][j] = (sum[i - 1][j] + dp[i][j]) % mod;
      s[i][j] = ((dp[i][j] + s[i - 1][j] % mod) + sum[i - 1][j] % mod);
    }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++) {
      if (i > 1)
        (ans += 1LL * (((dp[j][i - 1] - dp[j][i - 2]) % mod + mod) % mod) *
                dp[j][n - i] % mod * (m - j + 1) % mod) %= mod;
      else
        (ans += 1LL * dp[j][i - 1] * dp[j][n - i] % mod * (m - j + 1) % mod);
    }
  cout << ans << '\n';
  return 0;
}
