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;
long long Dp[2010][2010], S[2010][2010], K[2010][2010], Q[2010][2010], n, m,
    ans = 0;
int main() {
  scanf("%lld%lld", &n, &m);
  for (int i = 0; i <= n; i++) Dp[2][i] = 1, Dp[3][i] = i * 2 + 1;
  for (int i = 2; i <= m; i++) Dp[i][0] = 1, S[i][0] = 1;
  for (int i = 4; i <= m; i++)
    for (int j = 1; j <= n; j++)
      Dp[i][j] = (Dp[i - 1][j] * 2 % 1000000007 - Dp[i - 2][j] + Dp[i][j - 1] +
                  1000000007) %
                 1000000007;
  for (int i = 2; i <= m; i++)
    for (int j = 1; j <= n; j++)
      S[i][j] = (S[i][j - 1] + Dp[i][j]) % 1000000007;
  for (int i = 2; i <= m; i++)
    for (int j = 0; j <= n; j++) K[i][j] = (K[i - 1][j] + S[i][j]) % 1000000007;
  for (int i = 2; i <= m; i++)
    for (int j = 0; j <= n; j++) Q[i][j] = (Q[i - 1][j] + K[i][j]) % 1000000007;
  for (int i = 2; i <= m; i++) {
    ans = (ans + (long long)S[i][n - 1] * (m - i + 1)) % 1000000007;
    for (int j = 0; j <= n - 2; j++)
      ans = (ans + (long long)S[i][j] *
                       (Q[i - 1][n - j - 2] + K[i - 1][n - j - 2] + 1) %
                       1000000007 * (m - i + 1)) %
            1000000007;
  }
  printf("%lld\n", (ans % 1000000007 + 1000000007) % 1000000007);
  return 0;
}
