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;
#pragma comment(linker, "/STACK:64000000")
const int maxn = 1 << 17;
const int inf = 1000000007;
const int mod = 1000000007;
int n, m;
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int slowsolve() {
  int res = 0;
  vector<vector<int> > dp(n + 1, vector<int>(m, 0));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j < m; j++) {
      dp[i][j] = 1;
      for (int k = 1; k <= j; k++) {
        add(dp[i][j], 1LL * dp[i - 1][k] * (j - k + 1) % mod);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j < m; j++) {
      int cur = 1;
      for (int k = 1; k < j; k++) {
        add(cur, 1LL * dp[n - i][k] * (j - k + 1) % mod);
      }
      add(res, 1LL * dp[i][j] * cur % mod * (m - j) % mod);
    }
  }
  return res;
}
int fastsolve() {
  int res = 0;
  vector<vector<int> > dp(n + 1, vector<int>(m, 0));
  vector<vector<int> > sum1(n + 1, vector<int>(m, 0));
  vector<vector<int> > sum2(n + 1, vector<int>(m, 0));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j < m; j++) {
      dp[i][j] = 1;
      add(dp[i][j], 1LL * (j + 1) * sum1[i - 1][j] % mod);
      add(dp[i][j], -sum2[i - 1][j]);
      sum1[i][j] = sum1[i][j - 1];
      add(sum1[i][j], dp[i][j]);
      sum2[i][j] = sum2[i][j - 1];
      add(sum2[i][j], 1LL * j * dp[i][j] % mod);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j < m; j++) {
      int cur = 1;
      add(cur, 1LL * (j + 1) * sum1[n - i][j] % mod);
      add(cur, mod - sum2[n - i][j]);
      add(cur, mod - dp[n - i][j]);
      add(res, 1LL * dp[i][j] * cur % mod * (m - j) % mod);
    }
  }
  return res;
}
int main() {
  while (cin >> n >> m) cout << fastsolve() << endl;
  return 0;
}
