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;
namespace zyt {
template <typename T>
inline void read(T &x) {
  char c;
  bool f = false;
  x = 0;
  do c = getchar();
  while (c != '-' && !isdigit(c));
  if (c == '-') f = true;
  do x = x * 10 + c - '0', c = getchar();
  while (isdigit(c));
  if (f) x = -x;
}
template <typename T>
inline void write(T x) {
  static char buf[20];
  char *pos = buf;
  if (x < 0) putchar('-'), x = -x;
  do *pos++ = x % 10 + '0';
  while (x /= 10);
  while (pos > buf) putchar(*--pos);
}
const int N = 2e3 + 10, Jumpmelon = 1000000007;
int n, m, f[N][N];
int g[N][N];
int sum[N][N];
int h[N][N], sumgi[N], sumh[N][N], sh[N][N];
int work() {
  read(n), read(m);
  if (m < 2) {
    write(0);
    return 0;
  }
  for (int i = 2; i <= m; i++)
    f[1][i] = 1, sum[1][i] = sum[1][i - 1] + 1, h[0][i] = 1;
  for (int i = 2; i <= n; i++)
    for (int j = 2; j <= m; j++) {
      f[i][j] = (f[i][j - 1] + sum[i - 1][j]) % Jumpmelon;
      sum[i][j] = (sum[i][j - 1] + f[i][j]) % Jumpmelon;
      h[i - 1][j] = (f[i][j] - f[i - 1][j] + Jumpmelon) % Jumpmelon;
    }
  for (int j = 2; j <= m; j++) sumh[0][j] = h[0][j];
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++)
      sumh[i][j] = (sumh[i - 1][j] + h[i][j]) % Jumpmelon;
  for (int j = 2; j <= m; j++) sh[0][j] = sumh[0][j];
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++)
      sh[i][j] = (sh[i - 1][j] + sumh[i][j]) % Jumpmelon;
  int ans = 0;
  for (int j = 2; j <= m; j++) {
    sumgi[j] = 0;
    for (int i = 1; i <= n; i++)
      sumgi[j] = (sumgi[j] +
                  (long long)f[i][j] % Jumpmelon * sh[n - i][j] % Jumpmelon) %
                 Jumpmelon;
    ans = (ans + (long long)sumgi[j] * (m - j + 1) % Jumpmelon) % Jumpmelon;
  }
  write(ans);
  return 0;
}
}  // namespace zyt
int main() { return zyt::work(); }
