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;
int n, m;
const int mo = 1e9 + 7;
int dp[2005][2005], s1[2005][2005], s2[2005][2005];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    for (int j = 2; j <= m; ++j) {
      dp[i][j] = (1 + 1ll * s1[i - 1][j] * (j + 1) % mo - s2[i - 1][j]) % mo;
      s1[i][j] = (s1[i][j - 1] + dp[i][j]) % mo;
      s2[i][j] = (s2[i][j - 1] + 1ll * dp[i][j] * j % mo) % mo;
    }
  }
  int ans = 0;
  for (int d = 1; d <= n; ++d) {
    for (int k = 2; k <= m; ++k) {
      ans = (ans + 1ll * (m - k + 1) * (dp[d][k] - dp[d - 1][k]) % mo *
                       dp[n - d + 1][k] % mo) %
            mo;
    }
  }
  if (ans < 0) ans += mo;
  printf("%d\n", ans);
  return 0;
}
