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 mod = 1000000007;
const int N = 2000, M = N;
int n, m;
int dp[N + 1][M + 1];
int Sum[N + 1][M + 1], Sumk[N + 1][M + 1];
int sum[N + 2];
int main() {
  cin >> n >> m;
  for (int i = 2; i <= m; i++)
    dp[1][i] = 1, Sum[1][i] = (Sum[1][i - 1] + dp[1][i]) % mod,
    Sumk[1][i] = (Sumk[1][i - 1] - 1ll * i * dp[1][i]) % mod;
  for (int i = 2; i <= n; i++)
    for (int j = 2; j <= m; j++) {
      dp[i][j] = (1ll * (j + 1) * Sum[i - 1][j] + Sumk[i - 1][j] + 1) % mod;
      Sum[i][j] = (Sum[i][j - 1] + dp[i][j]) % mod;
      Sumk[i][j] = (Sumk[i][j - 1] - 1ll * j * dp[i][j]) % mod;
    }
  int ans = 0;
  for (int k = 2; k <= m; k++) {
    for (int j = n; j; j--)
      sum[j] = (1ll * sum[j + 1] + dp[n - j + 1][k] - dp[n - j][k]) % mod;
    for (int i = 1; i <= n; i++)
      (ans += 1ll * (m - k + 1) * (dp[i][k] - dp[i - 1][k]) % mod * sum[i] %
              mod) %= mod;
  }
  cout << (ans + mod) % mod;
  return 0;
}
