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;
long long mod = 1000000007;
int i, j, n, m;
long long dp[2005][2005], a[2005][2005], b[2005][2005];
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) {
    for (j = 2; j <= m; j++) {
      dp[i][j] = 1 + a[i - 1][j] * j - b[i - 1][j];
      if (dp[i][j] >= mod) dp[i][j] %= mod;
    }
    for (j = 2; j <= m; j++) {
      a[i][j] += a[i][j - 1] + dp[i][j];
      if (a[i][j] >= mod) a[i][j] %= mod;
    }
    for (j = 2; j <= m; j++) {
      b[i][j] += b[i][j - 1] + dp[i][j] * (j - 1);
      if (b[i][j] >= mod) b[i][j] %= mod;
    }
  }
  long long ans = 0;
  for (i = 1; i <= n; i++) {
    for (j = 2; j <= m; j++) {
      long long temp = dp[i][j];
      temp *= (1 + a[n - i][j] * j - b[n - i][j] - dp[n - i][j]) % mod;
      if (temp >= mod || temp <= -mod) temp %= mod;
      ans += temp * (m - j + 1);
      if (ans <= -mod || ans >= mod) ans %= mod;
    }
  }
  ans = (ans % mod + mod) % mod;
  printf("%I64d\n", ans);
  return 0;
}
