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 dx[] = {0, 0, 1, -1, 1, -1, 1, -1}, dy[] = {-1, 1, 0, 0, 1, -1, -1, 1};
void run();
int main() {
  ios::sync_with_stdio(0);
  cout << fixed << setprecision(16);
  run();
}
const int mod = 1000000007;
int n, m;
int f[2020][2020];
int g[2020][2020];
int s1[2020][2020], s2[2020][2020], s3[2020][2020], s4[2020][2020];
void run() {
  cin >> n >> m;
  for (int k = (2); k < (2020); ++k) {
    f[1][k] = 1;
    for (int j = 2; j <= 2000; ++j) {
      s1[1][j] = (s1[1][j - 1] + f[1][j]) % mod;
      s2[1][j] = (s2[1][j - 1] + (long long)j * f[1][j]) % mod;
    }
  }
  for (int i = 2; i <= 2000; ++i) {
    for (int w = 2; w <= 2000; ++w)
      f[i][w] = ((long long)(w + 1) * s1[i - 1][w] - s2[i - 1][w] + mod) % mod;
    for (int j = 2; j <= 2000; ++j) {
      s1[i][j] = (s1[i][j - 1] + f[i][j]) % mod;
      s2[i][j] = (s2[i][j - 1] + (long long)j * f[i][j]) % mod;
    }
  }
  for (int w = 2; w <= 2000; ++w)
    for (int h = 1; h <= 2000; ++h) {
      s3[h][w] = (s3[h - 1][w] + f[h][w]) % mod;
      s4[h][w] = (s4[h - 1][w] + (long long)(w + 1) * s1[h][w - 1] -
                  s2[h][w - 1] + mod) %
                 mod;
    }
  int ans = 0;
  for (int t = 1; t <= n; ++t) {
    for (int w = 2; w <= m; ++w) {
      int a = s3[t][w];
      int b = s4[n - t][w];
      ans = (ans + (long long)(m - w + 1) * a % mod * (1 + b)) % mod;
    }
  }
  cout << ans << "\n";
}
