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, Ans = 0, F[2005][2005], G[2005][2005], S[2005][2005];
int main() {
  scanf("%d%d", &N, &M);
  memset(F, 0, sizeof(F));
  memset(G, 0, sizeof(G));
  memset(S, 0, sizeof(S));
  for (int i = 1; i <= N; i++)
    for (int j = 2; j <= M; j++) {
      F[i][j] = (S[i - 1][j] + 1) % 1000000007;
      G[i][j] = (G[i][j - 1] + F[i][j]) % 1000000007;
      S[i][j] = (S[i][j - 1] + G[i][j]) % 1000000007;
    }
  for (int i = 1; i <= N; i++)
    for (int j = 2; j <= M; j++)
      Ans = (Ans + 1ll * (1ll * (M - j + 1) * F[i][j]) % 1000000007 *
                       (F[N - i + 1][j] - F[N - i][j] + 1000000007) %
                       1000000007) %
            1000000007;
  printf("%d\n", Ans);
  return 0;
}
