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;
template <class T>
inline void chkmin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline void chkmax(T &a, T b) {
  if (a < b) a = b;
}
int cnt[2100][2100], sum[2100][2100];
int ucnt[2100][2100], dcnt[2100][2100];
int main() {
  int N, M, b, h, i, ans;
  scanf("%d%d", &N, &M);
  for (b = 2; b <= M; b++) cnt[b][1] = 1, sum[b][1] = b - 1;
  for (h = 1; h <= N; h++) cnt[2][h] = sum[2][h] = 1;
  for (b = 3; b <= M; b++)
    for (h = 2; h <= N; h++) {
      cnt[b][h] = (sum[b][h - 1] + cnt[b - 1][h]) % 1000000007;
      sum[b][h] = (cnt[b][h] + sum[b - 1][h]) % 1000000007;
    }
  for (h = 1; h <= N; h++)
    for (b = 1; b <= M; b++) {
      cnt[b][h] += cnt[b][h - 1];
      cnt[b][h] %= 1000000007;
    }
  for (b = 2; b <= M; b++)
    for (h = 1; h <= N; h++) {
      ucnt[b][h] = cnt[b][h];
      dcnt[b][h] = cnt[b][h] - cnt[b][h - 1];
      if (dcnt[b][h] < 0) dcnt[b][h] += 1000000007;
    }
  ans = 0;
  for (i = 1; i <= N; i++) {
    for (b = 2; b <= M; b++) {
      ans += 1LL * ucnt[b][i] * dcnt[b][N - i + 1] % 1000000007 * (M - b + 1) %
             1000000007;
      if (ans >= 1000000007) ans -= 1000000007;
    }
  }
  printf("%d\n", ans);
  return 0;
}
