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 10:11:23.200991 */#include<cstdio>
#define mod 1000000007
long long f[2001][2001],ans;
int n,m;
int main()
{
    int i,j;
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)
    {
        long long sum=0;
        for(j=2;j<=m;j++)
        {
            sum+=f[i-1][j];
            sum%=mod;
            f[i][j]=(f[i][j-1]+sum)%mod;
        }
        for(j=2;j<=m;j++)f[i][j]++;
    }
    for(i=1;i<=n;i++)
        for(j=2;j<=m;j++)
            ans=(f[i][j]*(f[n-i+1][j]-f[n-i][j])*(m-j+1)+ans)%mod;
    printf("%d\n",(int)ans);
    return 0;
}