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 N = 2002;
const int MD = 1000000007;
long long n, m, ans, f[N][N], g[N][N];
int main() {
  cin >> n >> m;
  for (long long i = 2; i <= m; ++i) g[1][i] = f[1][i] = 1;
  long long s1, s2;
  for (long long i = 2; i <= n; ++i) {
    s1 = s2 = 0;
    for (int j = 2; j <= m; ++j) {
      f[i][j] = ((s1 * (j + 1) % MD - s2) % MD + MD) % MD;
      f[i][j] = (f[i][j] + 1) % MD;
      g[i][j] = (g[i - 1][j] + f[i][j]) % MD;
      s1 = (s1 + g[i - 1][j]) % MD;
      s2 = (s2 + g[i - 1][j] * j) % MD;
    }
  }
  for (long long i = 1; i <= n; ++i)
    for (long long j = 2; j <= m; ++j)
      ans = (ans + (m - j + 1) * f[i][j] % MD * g[n - i + 1][j] % MD) % MD;
  cout << ans << endl;
}
