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
import static java.lang.System.in;
import static java.lang.System.out;

import java.util.Scanner;


public class D295 {
	static long MAX = 1000000007;
	public static Scanner sc=new Scanner(in);
	public void run(){
		// TODO Auto-generated method stub
		int n=sc.nextInt(), m = sc.nextInt();
		m--;
		
		long [][] dp= new long[2010][2010];
		for (int i=1; i<=n; i++){
			long sub = 0, sum =0;
			for (int j=1; j<=m; j++){
				sub+=dp[i-1][j];
				sum+=sub;
				dp[i][j]= (sum + 1)%MAX;
			}		
		}
		long ans=0;
		for (int i=1; i<=n; i++){
			for (int j=1; j<=m; j++){
				ans = ans+ ((dp[i][j]*dp[n-i+1][j])%MAX)*(m-j+1); 
			}
			if (i<n){
				for (int j=1; j<=m; j++)
					ans = ans - ((dp[i][j]*dp[n-i][j])%MAX)*(m-j+1); 
			}
			ans%=MAX;
		}
		ln((ans+MAX)%MAX);
	}
	
	public static void main(String[] args) {
		new D295().run();
	}
	public static void ln(Object obj){
		out.println(obj);
	}

}
