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 = 2e3 + 6.66;
const int MAXM = 2e3 + 6.66;
const long long MOD = 1e9 + 7;
long long dp[MAXN][MAXM];
long long dp1[MAXN][MAXM];
long long par[MAXN][MAXM];
long long par2[MAXN][MAXM];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int sz = 0; sz < m - 1; sz++) dp[0][sz] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < m - 1; j++) {
      if (i == 1) {
        dp[i][j] = 1;
        dp1[i][j] = 1;
        par[i][j] = j + 1;
        par2[i][j] = par2[i][j - 1] + j + 1;
        continue;
      }
      dp1[i][j] = (1 + ((2 + j) * par[i - 1][j - 1]) % MOD * 1LL -
                   par2[i - 1][j - 1] * 1LL + MOD) %
                  MOD;
      dp[i][j] =
          (1 + ((2 + j) * par[i - 1][j]) % MOD - par2[i - 1][j] + MOD) % MOD;
      par[i][j] = (par[i][j - 1] + dp[i][j]) % MOD;
      par2[i][j] = (par2[i][j - 1] + (1LL * (j + 1LL) * dp[i][j]) % MOD) % MOD;
    }
  }
  long long ans = 0LL;
  for (int t = 0; t < n; t++)
    for (int sz = 0; sz < m - 1; sz++) {
      ans = (ans +
             (((m - sz - 1LL) * 1LL * dp[t + 1][sz]) % MOD * dp1[n - t][sz]) %
                 MOD) %
            MOD;
    }
  cout << ans << endl;
  return 0;
}
