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>
#include<cstdio>
using namespace std;
const int N=4010,p=1e9+7;
int n,m,C[N][N],cs[N][N],f[N][N];
inline void A(int &x,int y){x+=y;if (x>=p) x-=p;}
int main(){
	for (int i=0;i<N;++i){
		C[i][0]=cs[i][0]=1;
		for (int j=1;j<=i;++j) A(C[i][j],C[i-1][j]+C[i-1][j-1]),A(cs[i][j],cs[i][j-1]+C[i][j]);
	}
	scanf("%d%d",&n,&m);
	f[0][0]=1;
	for (int i=0;i<=n;++i)
		for (int j=0;j<=m;++j){
			if (i) A(f[i][j],f[i-1][j]);
			if (j) A(f[i][j],cs[j-1][min(i,j-1)]);
		}
	int ans=0;
	for (int i=0;i<=n;++i)
		for (int j=0;j<=i;++j)
		A(ans,1ll*f[j][i-j]*C[m-1][i-j]%p);
	printf("%d",ans);
	return 0;
}