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;
long long dp[maxn][maxn];
long long s[maxn][maxn], ss[maxn][maxn];
void upd(long long &t, long long k) {
  t += k;
  t %= mod;
}
int main() {
  int n, i, j, m;
  long long t;
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    for (j = 0; j + 1 < m; j++) {
      long long t = 1 + s[i - 1][j] * (j + 1) - ss[i - 1][j];
      upd(dp[i][j], t);
    }
    s[i][0] = dp[i][0];
    for (j = 1; j + 1 < m; j++) {
      s[i][j] = s[i][j - 1] + dp[i][j], ss[i][j] = ss[i][j - 1] + dp[i][j] * j;
      s[i][j] %= mod;
      ss[i][j] %= mod;
    }
  }
  long long ans = 0;
  for (i = 1; i <= n; i++) {
    for (j = 0; j + 1 < m; j++) {
      long long t1 = dp[i][j] - dp[i - 1][j];
      long long t2 = dp[n - i + 1][j];
      ans += t1 * t2 % mod * (m - j - 1);
      ans %= mod;
    }
  }
  cout << (ans + mod) % mod << endl;
}
