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 long long MOD = 1e9 + 7;
long long n, m, dp[2005][2005], s1[2005][2005], s2[2005][2005], s[2005][2005],
    _s[2005][2005], __s[2005][2005];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 2; i <= m; i++) dp[1][i] = 1;
  for (int j = 2; j <= m; j++) {
    s1[1][j] = (s1[1][j - 1] + dp[1][j]) % MOD;
    s2[1][j] = (s2[1][j - 1] + dp[1][j] * j % MOD) % MOD;
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 2; j <= m; j++) {
      dp[i][j] = (-s2[i - 1][j] + s1[i - 1][j] * (j + 1) % MOD + MOD) % MOD;
      printf("%d %d %lld\n", i, j, dp[i][j]);
    }
    for (int j = 2; j <= m; j++) {
      s1[i][j] = (s1[i][j - 1] + dp[i][j]) % MOD;
      s2[i][j] = (s2[i][j - 1] + dp[i][j] * j % MOD) % MOD;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++) s[i][j] = (s[i - 1][j] + dp[i][j]) % MOD;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++) _s[i][j] = (_s[i][j - 1] + s[i][j]) % MOD;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++)
      __s[i][j] = (__s[i][j - 1] + s[i][j] * j % MOD) % MOD;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++) {
      ans = (ans + s[i][j] *
                       (_s[n - i][j - 1] * (j + 1) % MOD - __s[n - i][j - 1] +
                        MOD + 1) %
                       MOD * (m - j + 1) % MOD) %
            MOD;
    }
  printf("%lld\n", ans);
  return 0;
}
