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, ans, i, j, k, tmp, f[2005][2005];
int sum[2005][2005], h[2005][2005];
int main() {
  scanf("%d%d", &n, &m);
  for (i = 2; i <= m; i++) {
    f[i][1] = 1;
    sum[i][1] = i - 1;
  }
  for (i = 2; i <= n; i++)
    for (j = 2; j <= m; j++) {
      f[j][i] = (f[j - 1][i] + sum[j][i - 1]) % Mod;
      sum[j][i] = (f[j][i] + sum[j - 1][i]) % Mod;
    }
  memset(sum, 0, sizeof(sum));
  for (i = 1; i <= n; i++)
    for (j = m; j > 0; j--) {
      f[j][i] = (long long)f[j][i] * (m - j + 1) % Mod;
      sum[j][i] = (sum[j + 1][i] + f[j][i]) % Mod;
    }
  for (i = 2; i <= n; i++) {
    for (j = m; j > 1; j--) h[j][i] = (h[j + 1][i] + sum[j][i - 1]) % Mod;
    for (j = m; j > 1; j--) h[j][i] = (h[j][i] - f[j][i - 1] + Mod) % Mod;
  }
  memset(sum, 0, sizeof(sum));
  for (i = 2; i <= n; i++)
    for (j = m, tmp = 0; j > 1; j--) {
      tmp = (tmp + sum[j][i - 1]) % Mod;
      h[j][i] = (h[j][i] + tmp) % Mod;
      sum[j][i] = (sum[j + 1][i] + h[j][i]) % Mod;
    }
  for (i = 1; i <= m; i++)
    for (j = 1; j <= n; j++) {
      ans = (ans + (long long)f[i][j] * (n - j + 1) % Mod) % Mod;
      ans = (ans + (long long)h[i][j] * (n - j + 1) % Mod) % Mod;
    }
  printf("%d\n", ans);
  return 0;
}
