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
/* 2013-07-27 10:04:04.586605 */#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <ctime>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <cassert>
#include <bitset>

using namespace std;

const long long mod=1000000007;
long long f[2005][2005],g[2005][2005],t[2005][2005],s[2005][2005],ans;
int n,m,i,j,k;

int main() {
//	freopen("a.in","r",stdin);
//	freopen("a.out","w",stdout);
	scanf("%d%d",&n,&m);
	memset(f,0,sizeof(f));
	memset(s,0,sizeof(s));
	memset(g,0,sizeof(g));
	for (i=2; i<=m; ++i) {
		f[1][i]=1;
		s[1][i]=1;
	}
	for (i=2; i<=n; ++i) {
		for (j=2; j<=m; ++j) {
			g[i][j]=(g[i][j-1]+f[i-1][j])%mod;
			f[i][j]=(f[i][j-1]+g[i][j])%mod;
			s[i][j]=(s[i-1][j]+f[i][j]-f[i-1][j])%mod;
			if (s[i][j]<0) s[i][j]+=mod;
		}
	}
	for (i=1; i<=n; ++i) { 
		for (j=1; j<=m; ++j) {
			printf("%I64d ",s[i][j]);
		}
		printf("\n");
	}
	memset(f,0,sizeof(f));
	memset(t,0,sizeof(t));
	memset(g,0,sizeof(g));
	for (i=2; i<=m; ++i) {
		f[1][i]=1;
		t[1][i]=1;
	}
	for (i=2; i<=n; ++i) {
		for (j=2; j<=m; ++j) {
			g[i][j]=(g[i][j-1]+f[i-1][j])%mod;
			f[i][j]=(f[i][j-1]+g[i][j])%mod;
			t[i][j]=(t[i-1][j]+f[i][j])%mod;
		}
	}
	for (i=1; i<=n; ++i) {
		for (j=1; j<=m; ++j) {
			printf("%I64d ",t[i][j]);
		}
		printf("\n");
	}
	
	ans=0;
	for (i=1; i<=n; ++i) {
		for (j=2; j<=m; ++j) {
			ans+=s[i][j]*t[n-i+1][j]*(long long)(m-j+1);
			ans%=mod;
		}
	}
	printf("%I64d\n",ans);
	return 0;
}
