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;
const int N = 2000 + 5, M = 2000 + 5;
int n, m;
long long f[N][M], np_f[N][M], np_fm[N][M], pn_f[N][M], pp_f[N][M], pp_fm[N][M];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    if (i == 1)
      for (int j = 2; j <= m; ++j) f[i][j] = 1;
    else
      for (int j = 2; j <= m; ++j)
        f[i][j] = (np_f[i - 1][j] * (j + 1) - np_fm[i - 1][j] + mod) % mod;
    for (int j = 1; j <= m; ++j) {
      np_fm[i][j] = (np_fm[i][j - 1] + f[i][j] * j) % mod;
      np_f[i][j] = (np_f[i][j - 1] + f[i][j]) % mod;
      pn_f[i][j] = (pn_f[i - 1][j] + f[i][j]) % mod;
    }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) {
      pp_f[i][j] = (pp_f[i - 1][j] + np_f[i][j]) % mod;
      pp_fm[i][j] = (pp_fm[i - 1][j] + np_fm[i][j]) % mod;
    }
  long long ans = 0;
  for (int k = 1; k <= n; ++k)
    for (int j = 2; j <= m; ++j) {
      ans =
          (ans +
           (m - j + 1) * pn_f[k][j] % mod *
               ((1 + pp_f[n - k][j - 1] * (j + 1) - pp_fm[n - k][j - 1] + mod) %
                mod)) %
          mod;
    }
  printf("%I64d\n", ans);
  return 0;
}
