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 re() {
  long long ret = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f = (ch == '-') ? -f : f, ch = getchar();
  while (isdigit(ch)) ret = ret * 10 + ch - '0', ch = getchar();
  return ret * f;
}
const long long MOD = 1e9 + 7;
long long n, m, f[2005][2005], sum1[2005][2005], sum2[2005][2005],
    sum3[2005][2005], ans;
signed main() {
  n = re(), m = re();
  for (long long i = 1; i <= n; ++i)
    for (long long j = 2; j <= m; ++j)
      f[i][j] = (sum2[i - 1][j] + 1) % MOD,
      sum1[i][j] = (sum1[i][j - 1] + f[i][j]) % MOD,
      sum2[i][j] = (sum2[i][j - 1] + sum1[i][j]) % MOD;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 2; j <= m; ++j)
      ans +=
          (f[i][j] - f[i - 1][j]) * f[n - i + 1][j] % MOD * (m - j + 1) % MOD,
          ans %= MOD;
  printf("%lld\n", ans + (ans < 0) * MOD);
  return 0;
}
