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 NMax = 2100;
const long long INF = 1000000007;
int N, M;
long long dp[NMax][NMax], sum[NMax], sum1[NMax][NMax], ret, dp1[NMax][NMax];
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 2; i <= M; i++) dp[1][i] = 1;
  for (int i = 2; i <= N; i++) {
    long long tmp = 0;
    for (int j = 2; j <= M; j++) {
      tmp = (tmp + dp[i - 1][j]) % INF;
      sum[j] = (sum[j - 1] + (dp[i - 1][j] * (long long)(j - 1)) % INF);
      dp[i][j] = ((tmp * j) % INF - sum[j] + INF) % INF;
    }
  }
  for (int i = 2; i <= M; i++) sum1[i][1] = 1;
  for (int i = 2; i <= N; i++) {
    long long tmp = 0;
    sum1[2][i] = sum1[2][i - 1];
    for (int j = 2; j <= M; j++) {
      tmp = (tmp + dp[i - 1][j]) % INF;
      sum[j] = (sum[j - 1] + (dp[i - 1][j] * (j - 1)) % INF) % INF;
      sum1[j][i] =
          (sum1[j][i - 1] + ((tmp * j) % INF - sum[j] + INF) % INF) % INF;
    }
  }
  long long ret = 0;
  for (int i = 1; i <= N; i++) {
    long long tmp = 0;
    dp1[i][2] = sum1[2][N - i + 1];
    ret = (ret + (dp1[i][2] * (M - 1)) % INF) % INF;
    for (int j = 2; j < M; j++) {
      tmp = (tmp + sum1[j][i - 1]) % INF;
      sum[j] = (sum[j - 1] + (sum1[j][i - 1] * (j - 1)) % INF) % INF;
      dp1[i][j + 1] = ((tmp * (j + 1) % INF - sum[j] + INF) % INF *
                       sum1[j + 1][N - i + 1]) %
                      INF;
      dp1[i][j + 1] = (dp1[i][j + 1] + sum1[j + 1][N - i + 1]) % INF;
      ret = (ret + (dp1[i][j + 1] * (M - j)) % INF) % INF;
    }
  }
  printf("%lld\n", ret);
  getchar();
  getchar();
  return 0;
}
