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 syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<P,int> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

const int M=2005;
ll A,B,C[M][M],S[M][M+1],dp[2*M][M];

int main(){
	cin>>A>>B;
	C[0][0]=S[0][1]=dp[0][0]=1;
	for(int i=1;i<M;i++) for(int j=0;j<=i;j++) C[i][j]=(C[i-1][j]+(j?C[i-1][j-1]:0))%mod;
	for(int i=1;i<M;i++) for(int j=1;j<=i+1;j++) S[i][j]=(C[i][j-1]+S[i][j-1])%mod;
	for(int i=1;i<2*M;i++) for(int j=0;j<=min(M,i);j++){
		if(j) dp[i][j]=(dp[i-1][j]+S[j-1][j]-S[j-1][max(0,min(j,2*j-i-1))]+mod)%mod;
		else dp[i][j]++;
	}
	ll res=0;
	for(int i=0;i<=A;i++) for(int j=max(1ll,B-A+i);j<=B;j++) (res+=C[B-1][j-1]*dp[A-i][B-j])%=mod;
	cout<<res<<endl;
}