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 = 2020;
const int MOD = 1e9 + 7;
long long int memo[MAXN][MAXN];
long long int go(int w, int h) {
  if (w < 2) return 0;
  if (h == 0) return 1;
  long long int &res = memo[w][h];
  if (res != -1) return res;
  res = (go(w, h - 1) + 2 * go(w - 1, h) - go(w - 2, h)) % MOD;
  if (res < 0) res += MOD;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  fill(memo[0], memo[m + 2], -1);
  for (int w = 0; w < (int)(m + 1); w++)
    for (int h = 0; h < (int)(n + 1); h++) {
      memo[w][h] = go(w, h);
    }
  for (int w = 0; w < (int)(m + 1); w++)
    for (int h = 0; h < (int)(n + 1); h++) {
      memo[w][h + 1] = (memo[w][h + 1] + memo[w][h]) % MOD;
    }
  long long int res = 0;
  for (int i = 0; i < (int)(n); i++) {
    for (int w = (int)(2); w < ((int)m + 1); w++) {
      long long int up = go(w, i), down = go(w, n - 1 - i);
      if (i < n - 1) down -= go(w, n - 2 - i);
      res += up * down % MOD * (m + 1 - w) % MOD;
    }
  }
  res %= MOD;
  if (res < 0) res += MOD;
  cout << res << endl;
  return 0;
}
