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<bits/stdc++.h>
#define fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
typedef long long ll;
using namespace std;
const int maxn=4e3+5,mo=1e9+7;
int n,m,ji[maxn],ni[maxn];
int a,b,c,d,i,sum,ans,f[maxn];
int qs(int x,int y){
	int s=1;
	for(;y;y>>=1,x=(ll)x*x%mo) if (y&1) s=(ll)s*x%mo;
	return s;
}
int calc(int n,int m){
	return (ll)ji[n]*ni[m]%mo*ni[n-m]%mo;
}
int main(){
	scanf("%d%d",&n,&m);
	ji[0]=1;
	fo(i,1,n+m) ji[i]=(ll)ji[i-1]*i%mo;
	ni[n+m]=qs(ji[n+m],mo-2);
	fd(i,n+m,1) ni[i-1]=(ll)ni[i]*i%mo;
	fo(a,0,n)
	fo(b,0,n-a) f[a+b]++;
	ans=f[n];
	fo(i,1,n) f[i]+=f[i-1];
	fo(b,1,m-1){
		a=m-b;
		int s1=calc(m-1,a);
		fo(d,1,min(a,m-b)){
			c=a-d;
			sum=n-(a+c);
			if (sum>=0){
				int s2=calc(a-1,c);
				(ans+=(ll)f[sum]*s1%mo*s2%mo)%=mo;
			}
		}
	}printf("%d\n",ans);
	
}