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;
int MOD = 1000000007;
const int MAXN = 2005;
long long DP[MAXN][MAXN], DP0[MAXN][MAXN];
int main() {
  int N, M;
  scanf("%d%d", &N, &M);
  if (M == 1) {
    printf("0\n");
    return 0;
  }
  for (int i = 0; i < (M); ++i) DP[0][i] = 1;
  for (int i = (1); i < (N + 1); ++i) {
    DP0[i - 1][0] = DP[i - 1][0];
    for (int j = (1); j < (M + 5); ++j)
      DP0[i - 1][j] = (DP[i - 1][j] + DP0[i - 1][j - 1]) % MOD;
    DP[i][0] = i + 1;
    for (int j = (1); j < (M + 5); ++j)
      DP[i][j] = (DP[i][j - 1] + DP0[i - 1][j]) % MOD;
  }
  long long result = 0;
  for (int t = 0; t < (N); ++t) {
    for (int w = 0; w < (M - 1); ++w) {
      long long tmp0 = (M - 1 - w) * DP[t][w] % MOD;
      long long tmp1 = DP[N - 1 - t][w];
      if (N - 1 - t != 0) tmp1 -= DP[N - 1 - t - 1][w];
      result += tmp0 * tmp1 % MOD;
    }
  }
  result %= MOD;
  result += MOD;
  result %= MOD;
  printf("%d\n", (int)result);
}
