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;
using ll = long long;
constexpr ll MOD = 1e9 + 7;
ll n, m;
ll dp[2005][2005];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (ll i = 1; i <= m; ++i) dp[1][i] = 1;
  for (ll i = 2; i <= n; ++i) {
    ll sum = 0;
    dp[i][1] = 1;
    for (ll j = 2; j <= m; ++j) {
      sum += dp[i - 1][j];
      sum %= MOD;
      dp[i][j] = dp[i][j - 1] + sum;
      dp[i][j] %= MOD;
    }
  }
  ll ans = 0;
  for (ll i = 1; i <= n; ++i) {
    for (ll j = 2; j <= m; ++j) {
      ll tmp = 0;
      tmp = dp[i][j] - dp[i - 1][j] + MOD;
      tmp %= MOD;
      tmp *= dp[n - i + 1][j];
      tmp %= MOD;
      tmp *= (m - j + 1);
      tmp %= MOD;
      ans += tmp;
      ans %= MOD;
    }
  }
  cout << ans << endl;
}
