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 10:36:34.493943 */#include<cstring>
#include<cstdio>
int n, m;
long long modulo = 1000000007;
long long sigma, ans, f[2111][2111], g[2111][2111];
int main()
{
	scanf("%d%d", &n, &m);
	f[1][1] = 0;
	for(int i = 2; i <= m; i++) f[i][1] = 1;
	for(int i = 2; i <= n; i++)
	{
		sigma = 0;f[1][i] = 1;
		for(int j = 2; j <= m; j++)
		{
			sigma += f[j][i-1]; sigma %= modulo;
			f[j][i] = f[j-1][i] + sigma; f[j][i] %= modulo;
		}
	}
	g[1][1] = 0;
	for(int i = 2; i <= m; i++) g[i][1] = 1;
	for(int i = 2; i <= n; i++)
	{
		sigma = 0; g[1][i] = 0;
		for(int j = 2; j <= m; j++)
		{
			g[j][i] = f[j][i];
			g[j][i] -= f[j][i-1];
		}
	}
	ans = 0;
	for(int i = 1; i <= n; i++)
	{
		for(int j = 2; j <= m; j++)
		{
			ans = ans + (m + 1 - j) * f[j][i] %modulo * g[j][n-i+1] % modulo;
			ans %= modulo;
		}
	}
	printf("%I64d\n", ans);
	fclose(stdin);
	return 0;
}
