Problem: 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

Solution (in CPP):

Solution:
//by yjz
#include<bits/stdc++.h>
using namespace std;
#define FF first
#define SS second
#define PB push_back
#define MP make_pair
#define bged(v) (v).begin(),(v).end()
#define foreach(it,s) for(__typeof((s).begin()) it=(s).begin();it!=(s).end();it++)
typedef long long ll;
const int Imx=2147483647;
const ll Lbig=2e18;
const int mod=1e9+7;
//My i/o stream
struct fastio
{
	char s[100000];
	int it,len;
	fastio(){it=len=0;}
	inline char get()
	{
		if(it<len)return s[it++];it=0;
		len=fread(s,1,100000,stdin);
		if(len==0)return EOF;else return s[it++];
	}
	bool notend()
	{
		char c=get();
		while(c==' '||c=='\n')c=get();
		if(it>0)it--;
		return c!=EOF;
	}
}_buff;
#define geti(x) x=getnum()
#define getii(x,y) geti(x),geti(y)
#define getiii(x,y,z) getii(x,y),geti(z)
#define puti(x) putnum(x),putchar(' ')
#define putii(x,y) puti(x),puti(y)
#define putiii(x,y,z) putii(x,y),puti(z)
#define putsi(x) putnum(x),putchar('\n')
#define putsii(x,y) puti(x),putsi(y)
#define putsiii(x,y,z) putii(x,y),putsi(z)
inline ll getnum()
{
	ll r=0;bool ng=0;char c;c=_buff.get();
	while(c!='-'&&(c<'0'||c>'9'))c=_buff.get();
	if(c=='-')ng=1,c=_buff.get();
	while(c>='0'&&c<='9')r=r*10+c-'0',c=_buff.get();
	return ng?-r:r;
}
template<class T> inline void putnum(T x)
{
	if(x<0)putchar('-'),x=-x;
	register short a[20]={},sz=0;
	while(x)a[sz++]=x%10,x/=10;
	if(sz==0)putchar('0');
	for(int i=sz-1;i>=0;i--)putchar('0'+a[i]);
}
inline char getreal(){char c=_buff.get();while(c==' '||c=='\n')c=_buff.get();return c;}
int a,b;
ll C[2011][2011],Cs[2011][2011],Css[2011][2011];
ll f(int a,int b){return b==0?1:Css[b-1][a];}
ll f2(int a,int b)
{
//	cerr<<"f2:"<<a<<" "<<b<<endl;
	ll ret=0;
	for(int i=0;i<=min(a,b);i++)
	{
//		cerr<<C[b][i]<<":"<<f(a-i,i)<<endl;
		ret+=C[b][i]*f(a-i,i)%mod;
	}
//	cerr<<"ret="<<ret<<endl;
	return ret;
}
int main()
{
	for(int i=0;i<=2005;i++)
	{
		C[i][0]=C[i][i]=1;
		for(int j=1;j<i;j++)C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
		Cs[i][0]=1;
		for(int j=1;j<=2005;j++)Cs[i][j]=(C[i][j]+Cs[i][j-1])%mod;
		Css[i][0]=1;
		for(int j=1;j<=2005;j++)Css[i][j]=(Cs[i][j]+Css[i][j-1])%mod;
		
	}
	cin>>a>>b;
	ll ans=0;
	for(int i=0;i<=a;i++)
	{
		ans+=f2(a-i,b-1);
	}
	cout<<ans%mod;
	return 0;
}