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;
long long N, M, prime = 1000000007;
long long dp[2005][2005] = {0};
int main() {
  FILE *INPUT = stdin;
  fscanf(INPUT, "%d%d", &N, &M);
  int i1 = 0, i2 = 0, i3 = 0;
  long long val1 = 0, val2 = 0, val3 = 0;
  for (i1 = 1; i1 <= N; ++i1) {
    val1 = 0;
    val2 = 1;
    for (i2 = 2; i2 <= M; ++i2) {
      val1 = (val1 + dp[i1 - 1][i2]) % prime;
      val2 = (val1 + val2) % prime;
      dp[i1][i2] = val2;
    }
  }
  long long caves = 0;
  for (i1 = 2; i1 <= M; ++i1) {
    val1 = 0;
    val2 = 0;
    for (i2 = 1; i2 <= N; ++i2) {
      val1 = dp[i2][i1] - dp[i2 - 1][i1];
      val2 = (val2 + val1 * dp[N - i2 + 1][i1]) % prime;
    }
    val2 = (val2 * (M - i1 + 1)) % prime;
    caves = (caves + val2) % prime;
  }
  printf("%d\n", (prime + caves) % prime);
  return 0;
}
