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;
inline int rd() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
const int MAXN = 3010;
long long N, M, ans;
long long f[MAXN][MAXN];
const long long mod = 1e9 + 7;
int main() {
  N = rd(), M = rd();
  for (int i = 2; i <= M; ++i) f[1][i] = 1;
  for (int i = 2; i <= N; ++i) {
    long long sum = 0;
    f[i][1] = 1;
    for (int j = 2; j <= M; ++j) {
      (sum += f[i - 1][j]) %= mod;
      (f[i][j] = f[i][j - 1] + sum) % mod;
    }
  }
  for (int i = 1; i <= N; ++i)
    for (int j = 2; j <= M; ++j) {
      ans = (ans + ((f[i][j] - f[i - 1][j]) % mod + mod) * f[N - i + 1][j] %
                       mod * (M - j + 1)) %
            mod;
    }
  printf("%lld\n", (ans % mod + mod) % mod);
  return 0;
}
