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 MAXSIZE = 2001;
const long long BASE = 1000000007;
int n, m;
long long dp[MAXSIZE + 1][MAXSIZE + 1];
long long sum[MAXSIZE + 1][MAXSIZE + 1];
int main() {
  scanf("%d%d", &n, &m);
  for (int l = 1; l < m; l++) dp[1][l] = 1;
  for (int i = 2; i <= n; i++) {
    long long sum = 0;
    for (int j = 1; j < m; j++) {
      sum += (dp[i - 1][j]) % BASE;
      dp[i][j] = (dp[i][j - 1] + sum) % BASE;
    }
  }
  for (int j = 1; j < m; j++)
    for (int t = n - 1; t >= 1; t--)
      sum[n - t][j] = (sum[n - t - 1][j] + dp[n - t][j]) % BASE;
  long long ANS = 0;
  for (int t = 1; t <= n; t++) {
    for (int j = 1; j < m; j++) {
      long long count = sum[n - t][j] - dp[n - t + 1][j] * (n - t) +
                        (dp[n - t + 1][j] - dp[0][j]) * (n - t + 1);
      count %= BASE;
      count = (count * (m - j)) % BASE;
      ANS = (ANS + dp[t][j] * count) % BASE;
    }
  }
  printf("%I64d", ANS);
  return 0;
}
