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 MaxN = 2010;
const int MOD = 1000000007;
int n, m;
long long F[MaxN][MaxN], sF[MaxN][MaxN], tF[MaxN][MaxN], a[MaxN][MaxN],
    b[MaxN][MaxN], c[MaxN][MaxN], d[MaxN][MaxN];
long long ans;
int main() {
  cin >> n >> m;
  for (int i = 1; i < m; ++i) {
    F[1][i] = 1;
    sF[1][i] = (sF[1][i - 1] + F[1][i]) % MOD;
    tF[1][i] = (tF[1][i - 1] + F[1][i] * i) % MOD;
  }
  for (int i = 2; i <= n; ++i)
    for (int j = 1; j < m; ++j) {
      F[i][j] = ((j + 1) * sF[i - 1][j] - tF[i - 1][j]) % MOD;
      sF[i][j] = (sF[i][j - 1] + F[i][j]) % MOD;
      tF[i][j] = (tF[i][j - 1] + F[i][j] * j) % MOD;
    }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j < m; ++j) {
      a[i][j] = (a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1] + F[i][j]) % MOD;
      b[i][j] =
          (b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1] + F[i][j] * i) % MOD;
      c[i][j] =
          (c[i - 1][j] + c[i][j - 1] - c[i - 1][j - 1] + F[i][j] * j) % MOD;
      d[i][j] =
          (d[i - 1][j] + d[i][j - 1] - d[i - 1][j - 1] + F[i][j] * i * j) % MOD;
    }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) {
      int ii = n - i, jj = j - 1, ki = n - i + 1, kj = j + 1;
      ans =
          (ans +
           (a[ii][jj] * ki * kj - b[ii][jj] * kj - c[ii][jj] * ki + d[ii][jj]) %
               MOD * (F[i][j] * (m - j) % MOD) +
           F[i][j] * (n - i + 1) * (m - j)) %
          MOD;
    }
  cout << ans << endl;
  return 0;
}
