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 = 1000000007;
void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void sub(int& a, int b) { add(a, mod - b); }
int N, M;
int f[2005][2005];
int s[2005][2005];
int t[2005][2005];
int main(void) {
  cin >> N >> M;
  for (int i = (1), _b = (N); i <= _b; i++)
    for (int j = (2), _b = (M); j <= _b; j++) {
      f[i][j] = (1 + long long(j + 1) * s[i - 1][j] - t[i - 1][j] + mod) % mod;
      s[i][j] = (s[i][j - 1] + f[i][j]) % mod;
      t[i][j] = (t[i][j - 1] + (long long)j * f[i][j]) % mod;
    }
  int ans = 0;
  for (int i = (1), _b = (N); i <= _b; i++)
    for (int j = (2), _b = (M); j <= _b; j++)
      ans = (ans + 1ll * f[i][j] *
                       (1 + (long long)(j + 1) * s[N - i][j - 1] -
                        t[N - i][j - 1] + mod) %
                       mod * (M - j + 1) % mod) %
            mod;
  cout << ans << endl;
  return 0;
}
