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;
int N, M;
signed long long mo = 1000000007;
signed long long dp[2002][2002], dp2[2002][2002], dp3[2002][2002],
    dp4[2002][2002];
void solve() {
  int i, j, k, l, r, x, y, w;
  string s;
  cin >> N >> M;
  for (x = 2; x <= 2000; x++) dp[x][0] = dp2[x][0] = dp3[x][0] = dp4[x][0] = 1;
  for (y = 1; y <= 2000; y++) {
    signed long long tot = 0;
    for (x = 2; x <= 2000; x++) {
      tot += dp[x][y - 1];
      dp[x][y] = (dp[x - 1][y] + tot) % mo;
      dp2[x][y] = (dp[x][y] + mo - dp[x][y - 1]) % mo;
      dp3[x][y] = (dp3[x][y - 1] + dp2[x][y]) % mo;
      dp4[x][y] = (dp4[x][y - 1] + dp3[x][y]) % mo;
    }
  }
  signed long long ret = 0;
  for (w = 2; w <= M; w++) {
    signed long long tmp = 0;
    for (y = 0; y < N; y++) tmp += dp3[w][y] * dp4[w][N - 1 - y] % mo;
    ret += tmp * (M + 1 - w) % mo;
  }
  cout << ret % mo << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false);
  for (i = 0; i < argc - 1; i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < s.size(); i++) ungetc(s[s.size() - 1 - i], stdin);
  solve();
  return 0;
}
