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 maxn = 2000 + 100;
const long long Mod = 1e9 + 7;
long long dp[maxn][maxn], sum[maxn][maxn], ssum[maxn][maxn];
long long fun(long long x) { return ((x % Mod) + Mod) % Mod; }
int main() {
  ios_base::sync_with_stdio(false);
  long long n, m;
  cin >> n >> m;
  for (long long i = 2; i <= m; i++)
    dp[1][i] = i * (i - 1) / 2, sum[1][i] = dp[1][i];
  for (long long i = 2; i <= n; i++)
    for (long long j = 2; j <= m; j++) {
      dp[i][j] = dp[i - 1][j] + dp[i][j - 1] + dp[i][j - 1] - dp[i][j - 2];
      dp[i][j] = fun(dp[i][j]);
      sum[i][j] = sum[i - 1][j] + dp[i][j];
      sum[i][j] = fun(sum[i][j]);
    }
  long long ans = 0;
  for (long long t = 1; t <= n; t++) {
    for (long long x = 2; x <= m; x++) {
      ans += (m - x + 1) * fun((1 + sum[t - 1][x]) *
                               fun(1 + sum[n - t][x - 1] + sum[n - t][x - 1] -
                                   sum[n - t][x - 2]));
      ans = fun(ans);
    }
  }
  cout << ans << endl;
}
