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
/* 2013-07-26 09:32:20.312362 */#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <queue>
#define mo 1000000007

using namespace std;

long long n, m, i, j, now1, now2, now, Ans;
long long f[2005][2005], f2[2005][2005];

int main()
{
	cin >> n >> m;
	f[0][0] = 1;
	for (i = 1; i <= n; i++)
	{
		f[i][0] = 1;
		now1 = 0;
		now2 = 1;
		for (j = 2; j <= m; j++)
		{
			now1 += f[i - 1][j], now1 %= mo;
			now2 += now1, now2 %= mo;
			f[i][j] = now2;
			f2[i][j] = (f[i][j] - f[i - 1][j]) % mo;
		}
	}
	
	Ans = 0;
	for (i = 1; i <= n; i++)
		for (j = 2; j <= m; j++)
		{
			now = f2[i][j];
			now *= f[n - i + 1][j];
			now %= mo;
			now *= (m - j + 1);
			now %= mo;
			Ans += now;
			Ans %= mo;
		}
	
	while (Ans < 0)
		Ans += mo;
	cout << Ans << endl;
	return 0;
}
