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 long long MOD = 1E9 + 7;
template <class C>
void mini(C& a4, C b4) {
  a4 = min(a4, b4);
}
template <class C>
void maxi(C& a4, C b4) {
  a4 = max(a4, b4);
}
long long n, m, dp[2007][2007], res;
int main() {
  ios_base::sync_with_stdio(false);
  cout << setprecision(15) << fixed;
  cin >> n >> m;
  m--;
  if (m <= 0) {
    cout << 0 << endl;
    return 0;
  }
  for (int p = (1); p <= (2007 - 1); ++p) {
    long long sum = 0, pref = 0;
    for (int l = (1); l <= (2007 - 1); ++l) {
      pref += dp[p - 1][l];
      pref %= MOD;
      sum += pref;
      sum %= MOD;
      dp[p][l] = sum + 1;
      dp[p][l] %= MOD;
    }
  }
  for (int r = (1); r <= (n); ++r)
    for (int s = (1); s <= (m); ++s) {
      res += ((dp[r][s] * dp[n - r + 1][s]) % MOD) * (m - s + 1);
      res %= MOD;
      if (r < n) {
        res -= ((dp[r][s] * dp[n - r][s]) % MOD) * (m - s + 1);
        res %= MOD;
      }
    }
  res %= MOD;
  if (res < 0) res += MOD;
  cout << res << endl;
  return 0;
}
