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;
const int N = 2100;
int n, m, d[N][N], sp1[N][N], sp2[N][N], rez, d1[N][N], spp1[N][N], spp2[N][N];
int main() {
  int i, j;
  cin >> n >> m;
  for (i = 2; i < N; ++i) {
    d[1][i] = d1[1][i] = spp1[1][i] = spp2[1][i] = 1;
    sp1[1][i] = (sp1[1][i - 1] + d[1][i]) % MOD;
    sp2[1][i] = (1LL * sp2[1][i - 1] + 1LL * d[1][i] * (i - 1)) % MOD;
  }
  for (i = 2; i < N; ++i)
    for (j = 2; j < N; ++j) {
      d[i][j] = (1LL * sp1[i - 1][j] * j - sp2[i - 1][j]) % MOD;
      if (d[i][j] < 0) d[i][j] += MOD;
      sp1[i][j] = (sp1[i][j - 1] + d[i][j]) % MOD;
      sp2[i][j] = (1LL * sp2[i][j - 1] + 1LL * d[i][j] * (j - 1)) % MOD;
      d1[i][j] = d[i][j] - d[i - 1][j];
      if (d1[i][j] < 0) d1[i][j] += MOD;
      spp1[i][j] = (spp1[i - 1][j] + d1[i][j]) % MOD;
      spp2[i][j] = (1LL * spp2[i - 1][j] + 1LL * d1[i][j] * i) % MOD;
    }
  for (i = 0; i <= n; ++i)
    for (j = 2; j <= m; ++j) {
      int te = (1LL * (n - i + 1) * spp1[n - i][j] - spp2[n - i][j]) % MOD;
      int t = (1LL * d[i + 1][j] * te) % MOD;
      t = (1LL * t * (m - j + 1)) % MOD;
      rez = (rez + t) % MOD;
    }
  cout << rez;
  return 0;
}
