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 mod = 1000000007;
const int maxn = 2000 + 100;
long long dp[maxn][maxn];
long long sum[maxn][maxn];
int main() {
  int n, m;
  cin >> n >> m;
  for (int j = 1; j <= m; j++) dp[1][j] = j * (j - 1) / 2;
  for (int i = 2; i <= n; i++)
    for (int j = 1; j <= m; j++)
      dp[i][j] = (dp[i - 1][j] + dp[i][j - 1] * 2 - dp[i][j - 2]) % mod;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) sum[i][j] = (sum[i - 1][j] + dp[i][j]) % mod;
  long long ans = 0;
  for (int t = 1; t <= n; t++)
    for (int j = 2; j <= m; j++)
      ans = (ans + ((m - j + 1) % mod *
                    ((sum[t - 1][j] + 1) *
                     (1 + sum[n - t][j - 1] * 2 - sum[n - t][j - 2]) % mod) %
                    mod) %
                       mod) %
            mod;
  cout << (ans + mod) % mod << endl;
}
