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<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define llong long long
using namespace std;

const int N = 2000;
const int P = 1e9+7;
llong fact[N+N+3],finv[N+N+3];
llong comb[N+3][N+3],sum[N+3][N+3];
int a,b;

void initcomb()
{
	comb[0][0] = 1ll;
	for(int i=1; i<=N; i++)
	{
		comb[i][0] = comb[i][i] = 1ll;
		for(int j=1; j<i; j++)
		{
			comb[i][j] = (comb[i-1][j]+comb[i-1][j-1])%P;
		}
	}
	sum[0][0] = 1ll;
	for(int i=0; i<=N; i++)
	{
		sum[i][0] = 1ll;
		for(int j=1; j<=N; j++) sum[i][j] = (sum[i][j-1]+comb[i][j])%P;
	}
	for(int i=0; i<=N; i++)
	{
		for(int j=1; j<=N; j++) sum[i][j] = (sum[i][j]+sum[i][j-1])%P;
	}
}

int main()
{
	initcomb();
	scanf("%d%d",&a,&b);
	if(a==0 || b==0) {printf("1"); return 0;}
	llong ans = 0ll;
	for(int i=1; i<=a+1; i++)
	{
		llong tmp = 1ll;
		for(int j=1; j<=min(i-1,b-1); j++)
		{
			tmp = (tmp+sum[j-1][i-j-1]*comb[b-1][j])%P;
//			printf("ij%d %d %lld\n",i,j,sum[j-1][i-j-1]*comb[b-1][j]);
		}
//		printf("%d %lld\n",i,tmp);
		ans = (ans+tmp)%P;
	}
	printf("%lld\n",ans);
	return 0;
}