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 mod = 1e9 + 7;
int n, m;
int f[2005][2005];
int c[6005][6005];
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < (int)(6005); ++i)
    for (int j = 0; j < (int)(i + 1); ++j)
      c[i][j] = j ? (c[i - 1][j - 1] + c[i - 1][j]) % mod : 1;
  for (int l = 1; l <= n; ++l)
    for (int y = 2; y <= m; ++y)
      f[l][y] = (c[2 * l + y - 4][y - 2] + f[l - 1][y]) % mod;
  int ans = 0;
  for (int y = 2; y <= m; ++y) {
    (ans += 1LL * (m - y + 1) % mod * f[n][y] % mod) %= mod;
    for (int t = 2; t <= n; ++t)
      (ans +=
       1LL * (m - y + 1) % mod * f[t][y] % mod * f[n - t + 1][y] % mod) %= mod;
    for (int t = 2; t <= n; ++t)
      (ans += mod - 1LL * (m - y + 1) % mod * f[t - 1][y] % mod *
                        f[n - t + 1][y] % mod) %= mod;
  }
  printf("%d\n", ans);
  return 0;
}
