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 = 2000 + 100;
const int mod = 1000000000 + 7;
int n, m;
int C[3 * maxn][3 * maxn];
int dp[maxn][maxn];
inline int fix(int v) {
  if (v >= mod) v -= mod;
  return v;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0, _n = (int)(3 * maxn); i < _n; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = (int)(1), _n = (int)(i); j < _n; j++)
      C[i][j] = fix(C[i - 1][j] + C[i - 1][j - 1]);
  }
  for (int h = 0, _n = (int)(n); h < _n; h++)
    for (int w = (int)(1), _n = (int)(m); w < _n; w++)
      dp[h][w] = fix((h ? dp[h - 1][w] : 0) + (C[2 * h + w - 1][w - 1]));
  int ans = 0;
  for (int h = 0, _n = (int)(n); h < _n; h++)
    for (int w = (int)(1), _n = (int)(m); w < _n; w++)
      ans = fix(ans + (m - w) * (long long)(C[2 * h + w - 1][w - 1]) % mod *
                          (long long)dp[n - h - 1][w] % mod);
  cout << ans << endl;
  {
    int _;
    cin >> _;
    return 0;
  }
}
