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 = 1000000007;
long long dp[2010][2010];
long long sum[2010][2010];
long long f[2010][2010];
int main() {
  memset(dp, 0, sizeof(dp));
  memset(sum, 0, sizeof(sum));
  long long n, m;
  cin >> n >> m;
  for (long long i = 2; i <= m; i++) {
    dp[1][i] = 1;
    sum[1][i] = sum[1][i - 1] + 1;
  }
  for (long long i = 2; i <= n; i++) {
    for (long long j = 2; j <= m; j++) {
      if (j == 2)
        dp[i][j] = (1 + dp[i][j - 1] + sum[i - 1][j]) % mod;
      else
        dp[i][j] = (dp[i][j - 1] + sum[i - 1][j]) % mod;
      sum[i][j] = (sum[i][j - 1] + dp[i][j]) % mod;
    }
  }
  memset(f, 0, sizeof(f));
  memset(sum, 0, sizeof(sum));
  for (long long i = 1; i <= n; i++) {
    for (long long j = 3; j <= m; j++) {
      sum[i][j] = (sum[i][j - 1] + dp[i][j - 1]) % mod;
      f[i][j] = (f[i][j - 1] + sum[i][j] + dp[i][j - 1]) % mod;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= m; i++) dp[0][i] = dp[i][1] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 2; j <= m; j++) {
      long long tmp = 0;
      if (i != n && j != 2)
        tmp = 1LL * dp[i][j] * (m - j + 1) % mod * (1 + f[n - i][j]) % mod;
      else if (j == 2)
        tmp = 1LL * dp[i][j] * (m - j + 1) % mod;
      else
        tmp = 1LL * dp[i][j] * (m - j + 1) % mod;
      ans = (ans + tmp) % mod;
    }
  }
  cout << ans << endl;
  return 0;
}
