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>
const int mod = 1000000007;
long long f[2005][2005], g[2005][2005];
int main() {
  for (int n, m; scanf("%d%d", &n, &m) != EOF;) {
    memset(f, 0, sizeof(f));
    memset(g, 0, sizeof(g));
    for (int i = 1; i <= n; i++) {
      long long sum = 0;
      for (int j = 2; j <= m; j++) {
        f[i][j] = (g[i - 1][j] + 1) % mod;
        sum = (sum + f[i][j]) % mod;
        g[i][j] = (g[i][j - 1] + sum) % mod;
      }
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 2; j <= m; j++) {
        ans = ans + (f[i][j] * (g[n - i][j] - f[n - i][j] + 1) % mod) *
                        (m - j + 1) % mod;
        ans = ans % mod;
      }
    }
    printf("%I64d\n", ans);
  }
}
