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;
template <class T>
T inline Abs(const T& x) {
  return x < 0 ? -x : x;
}
template <class T>
T inline Sqr(const T& x) {
  return x * x;
}
long long mod = 1000000000 + 7;
inline long long Sum(long long a, long long b) { return (a + b + mod) % mod; }
inline long long Mul(long long a, long long b) { return (a * b) % mod; }
const int maxn = 2010;
long long a[maxn][maxn];
long long b[maxn][maxn];
int main(int argc, char* argv[]) {
  long long n, m;
  cin >> n >> m;
  for (int k = 2; k <= m + 1; ++k) {
    for (int l = 1; l <= n + 1; ++l) {
      if (k > 2) {
        a[k][l] = Sum(a[k - 1][l], b[k][l - 1]);
      } else {
        a[k][l] = Sum(1, a[k][l - 1]);
      }
      b[k][l] = b[k - 1][l] + a[k][l];
    }
  }
  long long ans = 0;
  for (int u = 1; u <= n; ++u) {
    int d = n - u;
    for (int k = 2; k <= m; ++k) {
      ans = Sum(ans, Mul(m - k + 1, a[k][u]));
      if (u < n) {
        ans = Sum(ans,
                  Mul(m - k + 1, Mul(a[k][u], Sum(a[k][d + 1], -a[k][d] - 1))));
      }
    }
  }
  cout << ans << endl;
  return 0;
}
