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;
  int ch = getchar(), f = 1;
  while (!isdigit(ch) && (ch != '-') && (ch != EOF)) ch = getchar();
  if (ch == '-') {
    f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int N = 2010, mod = 1e9 + 7;
int n, m;
int dp[N][N], sum[N][N], Sum[N][N];
inline void init() {
  n = read(), m = read();
  for (register int i = (0); i <= (m - 2); i++)
    dp[1][i] = 1, sum[1][i] = sum[1][i - 1] + 1, Sum[1][i] = Sum[1][i - 1] + i;
  for (register int i = (2); i <= (n); i++)
    for (register int j = (0); j <= (m - 2); j++) {
      dp[i][j] =
          (1ll * sum[i - 1][j] * (j + 1) % mod - Sum[i - 1][j] + mod) % mod;
      sum[i][j] = (sum[i][j - 1] + dp[i][j]) % mod,
      Sum[i][j] = (Sum[i][j - 1] + 1ll * dp[i][j] * j % mod) % mod;
    }
}
int ans, Ans[N][N];
inline void calc() {
  for (register int i = (1); i <= (n); i++)
    for (register int j = (0); j <= (m - 2); j++)
      Ans[i][j] = (Ans[i - 1][j] + dp[i][j]) % mod;
  for (register int i = (1); i <= (n); i++)
    for (register int j = (0); j <= (m - 2); j++)
      ans = (ans + 1ll * Ans[i][j] * (m - j - 1) % mod *
                       (Ans[n - i + 1][j] - Ans[n - i][j] + mod) % mod) %
            mod;
  printf("%d\n", ans);
}
int main() { init(), calc(); }
