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 11:20:20.524072 */#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define Mo 1000000007
using namespace std;
int F[2005][2005],S[2005][2005];
inline void Up(int &x,int y){
    x+=y;
    if(x>=Mo)x-=Mo;
}
int main(){
    int n,m,i,j,k,l;
    scanf("%d%d",&n,&m);
    for(i=0;i<=m-2;i++)F[1][i]=1;
    for(i=2;i<=n;i++){
        int s=0,s1=0;
        for(j=0;j<=m-2;j++){
            Up(s1,F[i-1][j]);
            Up(s,s1);
            Up(F[i][j],s);
        }
    } 
    int s1=0,s=0;
    for(j=0;j<=m-2;j++){
        s1=0;
        s=0;
        for(i=1;i<=n;i++){
            Up(s1,F[i][j]);
            Up(s,s1);
            Up(S[i][j],s);
        }
    }
/*for(i=1;i<=n;i++)
    for(j=0;j<=m-2;j++)printf("%d %d %d %d\n",i,j,F[i][j],S[i][j]);*/
    int ans=0;
    for(i=1;i<=n;i++){
        int s=0,s1=0;
        for(j=0;j<=m-2;j++){
            Up(ans,(((long long)F[i][j])*(n-i+1)%Mo*(m-j-1))%Mo);
            int tt=((long long)s)*(m-j-1)%Mo;
            tt=((long long)tt)*(F[i][j])%Mo;
            Up(ans,tt);
            Up(s1,S[n-i][j]);
            Up(s,(s1+S[n-i][j])%Mo);
        }
    }
    printf("%d\n",ans);
    return 0;
}
