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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
inline long long read() {
  long long t = 0, dp = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') dp = -1;
    c = getchar();
  }
  while (isdigit(c)) t = t * 10 + c - 48, c = getchar();
  return t * dp;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + 48);
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
inline void write_p(long long x) {
  write(x);
  putchar(' ');
}
const int N = 2005;
long long mo = 1e9 + 7, n, m, dp[N][N], s[N][N], sum[N][N], ans;
int main() {
  n = read();
  m = read();
  for (int i = 1; i <= n; ++i)
    for (int j = 2; j <= m; ++j) {
      dp[i][j] = (j + 1) * sum[i - 1][j] - s[i - 1][j] + 1;
      s[i][j] = s[i][j - 1] + dp[i][j] * j % mo;
      sum[i][j] = sum[i][j - 1] + dp[i][j];
      dp[i][j] = (dp[i][j] % mo + mo) % mo;
      s[i][j] %= mo;
      sum[i][j] %= mo;
    }
  for (int i = 1; i <= n; ++i)
    for (int j = 2; j <= m; ++j) {
      ans = ans + (dp[i][j] * (dp[n - i + 1][j] - dp[n - i][j] + mo) % mo *
                   (m - j + 1) % mo) %
                      mo;
      ans %= mo;
    }
  writeln(ans);
}
