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
//start:15:12
#include <bits/stdc++.h>
using namespace std;
#define int long long
int n,m;
const int N=2005;
const int mod=1e9+7;
int dp[N][N];
void add(int &x,int y){
	x=(x+y)%mod;
} 
signed main(){
	cin >> n >> m;
	for (int i=1;i<=m;i++) dp[1][i]=1;
	for (int i=2;i<=n;i++){
		dp[i][1]=1;int s=0;
		for (int j=2;j<=m;j++){
			(s+=dp[i-1][j])%mod;(dp[i][j]+=dp[i][j-1]+s)%=mod;
		}
	}
//	for (int i=1;i<=n;i++){
//		for (int j=1;j<=m;j++) cout << dp[i][j] << ' ';cout << endl;
//	}
	int ans=0;
	for (int i=1;i<=n;i++) for (int j=2;j<=m;j++){
		(ans+=(dp[i][j]-dp[i-1][j])*dp[n+1-i][j]%mod*(m-j+1))%=mod;
	}
	cout << ans << endl;
	return 0;
}