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;
const int MOD = 1000000007;
int add(int a, int b) { return a + b >= MOD ? a + b - MOD : a + b; }
int mul(int a, int b) { return (long long)a * b % MOD; }
int sub(int a, int b) { return a < b ? a - b + MOD : a - b; }
const int MAXN = 2001;
int n, m, f[MAXN][MAXN], g[MAXN][MAXN];
int main() {
  scanf("%d%d", &n, &m);
  for (int row = (1); row < (n + 1); row++) {
    int sumf = 0;
    for (int column = (2); column < (m + 1); column++) {
      f[row][column] = add(g[row - 1][column], 1);
      sumf = add(sumf, f[row][column]);
      g[row][column] = add(g[row][column - 1], sumf);
    }
  }
  int res = 0;
  for (int row = (1); row < (n + 1); row++) {
    for (int column = (2); column < (m + 1); column++) {
      res =
          add(res, mul(mul(f[row][column],
                           add(sub(g[n - row][column], f[n - row][column]), 1)),
                       m - column + 1));
    }
  }
  printf("%d\n", res);
  return 0;
}
