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 long long read() {
  long long x = 0;
  char ch = getchar(), w = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * w;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
const int N = 110000;
const long long p = 1e9 + 7;
int n, m;
long long f[2100][2100], ans;
int main() {
  n = read();
  m = read();
  f[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    f[i][0] = 1;
    long long sum = 1, res = 0;
    for (int j = 2; j <= m; ++j) {
      res += f[i - 1][j];
      res %= p;
      sum += res;
      sum %= p;
      f[i][j] = sum;
    }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 2; j <= m; ++j)
      (ans += (f[i][j] - f[i - 1][j]) * f[n - i + 1][j] % p * (m - j + 1)) %= p;
  ans = (ans + p) % p;
  writeln(ans);
}
