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 maxn = 2010, mod = 1000 * 1000 * 1000 + 7;
long long ans, dp[maxn][maxn], sum[maxn], n, m, t[maxn];
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    dp[2][i] = i;
    sum[i + 1] += i;
    t[i + 1] += i;
  }
  for (int i = 3; i <= m; i++) {
    for (int j = 1; j <= n; j++) {
      sum[j] += dp[i][j - 1];
      sum[j] %= mod;
      t[j] += sum[j];
      t[j] %= mod;
      dp[i][j] = (t[j] + 1) % mod;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 2; j <= m; j++) {
      long long x = (dp[j][i] * (dp[j][n - i + 1] - dp[j][n - i] + mod)) % mod;
      x = (x * (m + 1 - j)) % mod;
      ans += x;
      ans %= mod;
    }
  }
  cout << ans << endl;
  return 0;
}
