A + B balls are arranged in a row. The leftmost A balls are colored red, and the rightmost B balls are colored blue.

You perform the following operation:

* First, you choose two integers s, t such that 1 \leq s, t \leq A + B.
* Then, you repeat the following step A + B times: In each step, you remove the first ball or the s-th ball (if it exists) or the t-th ball (if it exists, all indices are 1-based) from left in the row, and give it to Snuke.



In how many ways can you give the balls to Snuke? Compute the answer modulo 10^9 + 7.

Here, we consider two ways to be different if for some k, the k-th ball given to Snuke has different colors. In particular, the choice of s, t doesn't matter. Also, we don't distinguish two balls of the same color.

Constraints

* 1 \leq A, B \leq 2000

Input

Input is given from Standard Input in the following format:


A B


Output

Print the answer.

Examples

Input

3 3


Output

20


Input

4 4


Output

67


Input

7 9


Output

7772


Input

1987 1789


Output

456315553
#include <iostream>
#define MOD 1000000007
using namespace std;
int C[2005][2005];
int pre[2005][2005];
int mat[2005][2005];
int A,B;
int add(int a,int b)
{
    a+=b;
    if(a>=MOD)a-=MOD;
    return a;
}
int mult(int a,int b)
{
    return 1LL*a*b%MOD;
}
int main()
{
    for(int i=0;i<=2000;i++)
    {
        C[i][0]=C[i][i]=1;
        for(int j=1;j<i;j++)
            C[i][j]=add(C[i-1][j],C[i-1][j-1]);
    }
    for(int i=0;i<=2000;i++)
    {
        pre[i][0]=C[i][0];
        for(int j=1;j<=i;j++)
            pre[i][j]=add(pre[i][j-1],C[i][j]);
    }
    for(int i=0;i<=2000;i++)
        mat[i][0]=mat[0][i]=1;
    for(int i=1;i<=2000;i++)
    {
        for(int j=1;j<=2000;j++)
        {
            mat[i][j]=add(mat[i-1][j],pre[j-1][min(i,j-1)]);
        }
    }
    cin>>A>>B;
    int ans=0;
    for(int i=0;i<=A;i++)
    {
        for(int j=0;j<=min(A-i,B-1);j++)
        {
            ans=add(ans,mult(mat[i][j],C[B-1][j]));
        }
    }
    cout<<ans;
    return 0;
}
