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:
#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){return o<<"("<<p.fs<<","<<p.sc<<")";}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){o<<"sz = "<<vc.size()<<endl<<"[";for(const T& v:vc) o<<v<<",";o<<"]";return o;}

template<unsigned int mod_>
struct ModInt{
	using uint = unsigned int;
	using ll = long long;
	using ull = unsigned long long;

	constexpr static uint mod = mod_;

	uint v;
	ModInt():v(0){}
	ModInt(ll v):v(normS(v%mod+mod)){}
	explicit operator bool() const {return v!=0;}
	static uint normS(const uint &x){return (x<mod)?x:x-mod;}		// [0 , 2*mod-1] -> [0 , mod-1]
	static ModInt make(const uint &x){ModInt m; m.v=x; return m;}
	ModInt operator+(const ModInt& b) const { return make(normS(v+b.v));}
	ModInt operator-(const ModInt& b) const { return make(normS(v+mod-b.v));}
	ModInt operator-() const { return make(normS(mod-v)); }
	ModInt operator*(const ModInt& b) const { return make((ull)v*b.v%mod);}
	ModInt operator/(const ModInt& b) const { return *this*b.inv();}
	ModInt& operator+=(const ModInt& b){ return *this=*this+b;}
	ModInt& operator-=(const ModInt& b){ return *this=*this-b;}
	ModInt& operator*=(const ModInt& b){ return *this=*this*b;}
	ModInt& operator/=(const ModInt& b){ return *this=*this/b;}
	ll extgcd(ll a,ll b,ll &x,ll &y) const{
		ll u[]={a,1,0},v[]={b,0,1};
		while(*v){
			ll t=*u/ *v;
			rep(i,3) swap(u[i]-=t*v[i],v[i]);
		}
		if(u[0]<0) rep(i,3) u[i]=-u[i];
		x=u[1],y=u[2];
		return u[0];
	}
	ModInt inv() const{
		ll x,y;
		extgcd(v,mod,x,y);
		return make(normS(x+mod));
	}
	bool operator==(const ModInt& b) const { return v==b.v;}
	bool operator!=(const ModInt& b) const { return v!=b.v;}
	friend istream& operator>>(istream &o,ModInt& x){
		ll tmp;
		o>>tmp;
		x=ModInt(tmp);
		return o;
	}
	friend ostream& operator<<(ostream &o,const ModInt& x){ return o<<x.v;}
};
using mint = ModInt<1000000007>;

mint C[2010][2010];
mint f[2010][2010];

mint solve1(int A,int B){
	if(A==0 || B==0) return 1;
	return f[B-1][A];
}

int main(){
	rep(i,2010){
		C[i][0] = C[i][i] = 1;
		rep1(j,i-1) C[i][j] = C[i-1][j-1] + C[i-1][j];
	}
	int A,B;
	cin>>A>>B;

	rep(X,B){
		mint now = 0;
		rep(y,A+2){
			if(X>=y) now += C[X][y];
			f[X][y] = (y==0?0:f[X][y-1]) + now;
		}
	}

	mint ans = 0;
	rep(P,A+1){
		rep(x,A-P+1){
			if(x>B-1) break;
			ans += C[B-1][x] * solve1(A-P-x,x);
		}
	}
	cout<<ans<<endl;
}