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 long long mod = 1e9 + 7;
long long n, m, sum, ans, f[2010][2010];
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ '0');
    c = getchar();
  }
  return x * f;
}
int main() {
  n = read();
  m = read();
  for (int i = 1; i <= m; i++) f[1][i] = 1;
  for (int i = 2; i <= n; i++) {
    sum = 0;
    f[i][1] = 1;
    for (int j = 2; j <= m; j++) {
      sum = (sum + f[i - 1][j]) % mod;
      f[i][j] = (f[i][j - 1] + sum) % mod;
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++)
      ans = (ans + (f[i][j] - f[i - 1][j]) * f[n - i + 1][j] % mod *
                       (m - j + 1) % mod) %
            mod;
  printf("%lld", (ans + mod) % mod);
  return 0;
}
