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 = 2000 + 5;
const long long mod = 1000 * 1000 * 1000 + 7;
long long d[maxN][maxN];
long long b[maxN][maxN];
int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  if (m < 2) {
    cout << 0 << endl;
    return 0;
  }
  for (int j = 0; j <= m - 2; j++) b[1][j] = d[1][j] = 1;
  for (int i = 2; i <= n; i++) {
    long long sum = 0;
    for (int j = 0; j <= m - 2; j++) {
      sum += d[i - 1][j];
      sum %= mod;
      d[i][j] = (sum + d[i][j - 1]) % mod;
      b[i][j] += d[i][j] + b[i - 1][j];
      b[i][j] %= mod;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= m - 2; j++)
      ans += ((((b[i][j] - b[i - 1][j]) * (b[n - i + 1][j])) % mod) *
              (m - 1 - j)) %
             mod,
          ans %= mod;
  if (ans < 0) ans += mod;
  cout << ans << endl;
  return 0;
}
