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>
const int MAXN = 2e3 + 10;
const long long MOD = 1e9 + 7;
using namespace std;
inline long long in() {
  long long x = 0, flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * flag;
}
int n, m;
long long ans;
long long up[MAXN][MAXN], dw[MAXN][MAXN];
int main() {
  n = in(), m = in();
  for (int i = 2; i <= m; i++) up[1][i] = 1;
  for (int i = 2; i <= n; i++) {
    long long presum = 0;
    up[i][1] = 1;
    for (int j = 2; j <= m; j++) {
      (presum += up[i - 1][j]) %= MOD;
      up[i][j] = (up[i][j - 1] + presum) % MOD;
    }
  }
  for (int i = 2; i <= m; i++) dw[n][i] = 1;
  for (int i = n - 1; i >= 1; i--) {
    long long presum = 0;
    dw[i][1] = 1;
    for (int j = 2; j <= m; j++) {
      if (j == 2)
        dw[i][j] = 1;
      else
        dw[i][j] = (dw[i][j - 1] + up[n - i][j - 1] + presum) % MOD;
      (presum += up[n - i][j]) %= MOD;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 2; j <= m; j++) {
      (ans += up[i][j] * dw[i][j] % MOD * (m - j + 1) % MOD) %= MOD;
    }
  }
  cout << ans;
  return 0;
}
