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>
#define mod 1000000007
#define int long long
using namespace std;
int n,m,dp[3000][3000];

main()
{
	cin>>n>>m;
	for(int i=1;i<=m;i++) dp[1][i]=1;
	for(int i=2;i<=n;i++)
	{
		int s=0;dp[i][1]=1;
		for(int j=2;j<=m;j++)
		{
			s+=dp[i-1][j],s%=mod;
			dp[i][j]=(dp[i][j-1]+s)%mod;
		}
	}	
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=2;j<=m;j++)
			ans=(ans+(dp[i][j]-dp[i-1][j])%mod*(dp[n-i+1][j])%mod*(m-j+1))%mod;
	cout<<ans%mod<<endl;
	return 0;
}
 	 	 			 	   	 		     			 	 			