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, P = 1e9 + 7, f[2010][2010], s1[2010][2010], s2[2010][2010],
          g[2010][2010];
void read(int &x) {
  char ch = getchar();
  int mark = 1;
  for (; ch != '-' && (ch < '0' || ch > '9'); ch = getchar())
    ;
  if (ch == '-') mark = -1, ch = getchar();
  for (x = 0; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - 48;
  x *= mark;
}
int main() {
  read(n);
  read(m);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      s1[i][j] = (s1[i][j - 1] + f[i - 1][j]) % P;
      s2[i][j] = (s2[i][j - 1] + (long long)f[i - 1][j] * j % P) % P;
    }
    for (int j = 2; j <= m; j++)
      f[i][j] = ((long long)s1[i][j] * (j + 1) % P - s2[i][j] + 1) % P;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      s1[i][j] = (s1[i][j - 1] + f[i][j]) % P;
      s2[i][j] = (s2[i][j - 1] + (long long)f[i][j] * j % P) % P;
      g[i][j] = ((long long)s1[i][j - 1] * (j + 1) % P - s2[i][j - 1]) % P;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      (ans += (long long)((long long)f[i][j] * (g[n - i][j] + 1) % P) *
              (m - j + 1) % P) %
          P;
  ans = (ans + P) % P;
  printf("%d\n", ans);
  return 0;
}
