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 Get() {
  int res = 0, q = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') q = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') res = res * 10 + ch - '0', ch = getchar();
  return res * q;
}
const double eps = 1e-12, pi = M_PI;
const int oo = (int)2e9, mod = (int)1e9 + 7;
const long long INF = (long long)1e17;
const int N = 2000;
int n, m, ans;
int f[N + 10][N + 10];
int main() {
  scanf(
      "%d"
      "%d"
      "\n",
      &n, &m);
  for (int i = (2), end = (m); i <= end; ++i) f[1][i] = 1;
  for (int i = (2), end = (n); i <= end; ++i) {
    int s = 0;
    for (int j = (2), end = (m); j <= end; ++j) {
      s = (s + f[i - 1][j]) % mod;
      f[i][j] = (f[i][j - 1] + s) % mod;
    }
  }
  for (int i = (1), end = (n); i <= end; ++i) {
    for (int j = (2), end = (m); j <= end; ++j) {
      f[i][j] = (f[i][j] + f[i - 1][j]) % mod;
    }
  }
  for (int i = (1), end = (n); i <= end; ++i) {
    long long cy = 1, s = 0;
    for (int j = (2), end = (m); j <= end; ++j) {
      long long cx = 1;
      cx = (cx + f[i][j] - f[1][j]) % mod;
      s = (s + f[n - i][j - 2]) % mod;
      cy = (cy + s) % mod, cy = (cy + f[n - i][j - 1]) % mod,
      cy = (cy + f[n - i][j - 1]) % mod;
      cx = cx * (long long)(m - j + 1) % mod;
      cx = cx * cy % mod;
      ans = (ans + cx) % mod;
    }
  }
  printf(
      "%d"
      "\n",
      ans);
  return 0;
}
