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 = 2010;
const int MAXM = 2010;
const long long MOD = int(1e+9) + 7;
int N, M;
long long dp[MAXN][MAXM], prefix1[MAXN][MAXM], prefix2[MAXN][MAXM];
int main() {
  scanf("%d%d", &N, &M);
  int i, j;
  for (i = 1; i <= M - 1; i++) {
    dp[0][i] = 0;
    dp[1][i] = 1;
  }
  for (i = 2; i <= N; i++) {
    prefix1[i - 1][0] = 0;
    for (j = 1; j <= M - 1; j++)
      prefix1[i - 1][j] = (prefix1[i - 1][j - 1] + dp[i - 1][j]) % MOD;
    prefix2[i - 1][0] = 0;
    for (j = 1; j <= M - 1; j++)
      prefix2[i - 1][j] = (prefix2[i - 1][j - 1] + prefix1[i - 1][j]) % MOD;
    for (j = 1; j <= M - 1; j++) dp[i][j] = (prefix2[i - 1][j] + 1) % MOD;
  }
  long long ans = 0;
  for (i = 1; i <= N; i++)
    for (j = 1; j <= M - 1; j++) {
      long long Q = dp[i][j] * (dp[N + 1 - i][j] - dp[N - i][j] + MOD) % MOD;
      ans = (ans + Q * (M - j)) % MOD;
    }
  cout << ans << endl;
}
