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 dp[2010][2010], g[2010][2010], h[2010][2010];
long long sum[2010];
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  for (int L = 2; L <= m; L++) {
    dp[1][L] = 1;
    sum[L] = (sum[L - 1] + dp[1][L]) % 1000000007;
  }
  for (int i = 2; i <= n; i++) {
    for (int L = 2; L <= m; L++)
      dp[i][L] = (dp[i][L - 1] + sum[L]) % 1000000007;
    for (int L = 2; L <= m; L++) sum[L] = (sum[L - 1] + dp[i][L]) % 1000000007;
  }
  for (int i = 1; i <= n; i++)
    for (int L = 2; L <= m; L++) {
      h[i][L] = h[i - 1][L] + dp[i][L];
      h[i][L] %= 1000000007;
    }
  for (int L = 2; L <= m; L++) g[0][L] = 1;
  for (int i = 1; i <= n; i++) {
    for (int L = 2; L <= m; L++) sum[L] = (dp[i][L] + sum[L - 1]) % 1000000007;
    long long tmp = 0;
    for (int L = 2; L <= m; L++) {
      tmp += sum[L];
      tmp %= 1000000007;
      g[i][L] = (g[i - 1][L] + tmp - dp[i][L] + 1000000007) % 1000000007;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++)
    for (int L = 2; L <= m; L++) {
      ans += h[i][L] * (m - L + 1) % 1000000007 * g[n - i][L] % 1000000007;
      ans %= 1000000007;
    }
  printf("%lld\n", ans);
  return 0;
}
