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 N = 2005;
const int MOD = 1e9 + 7;
int dp[N][N], n, m, gt[N][N][2];
inline int f(int w, int h) { return dp[w][h]; }
void do_dp2() {
  for (int i = 1; i < N; i++) {
    dp[i][1] = 1;
    dp[i][0] = 0;
    dp[2][i] = 1;
  }
  for (int j = 2; j < N; j++) {
    int sum = dp[2][j - 1], reg = dp[2][j - 1];
    for (int i = 3; i < N; i++) {
      reg = (reg + dp[i][j - 1]) % MOD;
      sum = (sum + reg) % MOD;
      dp[i][j] = sum;
    }
  }
}
int get(int x, int y, int lim) {
  return (f(x, y) - lim * f(x, y - 1) + MOD) % MOD;
}
int main() {
  memset(dp, -1, sizeof(dp));
  do_dp2();
  scanf("%d%d", &n, &m);
  long long int sol = 0;
  for (int i = 1; i < N; i++) {
    for (int j = 1; j < N; j++) {
      gt[i][j][0] = (get(i, j, 0) + gt[i][j - 1][0]) % MOD;
      gt[i][j][1] = (get(i, j, 1) + gt[i][j - 1][1]) % MOD;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int d = 2; d <= m; d++) {
      long long int sum1 = gt[d][i + 1][1], sum2 = gt[d][n - i][0];
      sol = (sum1 * sum2 % MOD * (m - d + 1) % MOD + sol) % MOD;
    }
  }
  printf("%d\n", sol);
}
