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 reg register int
#define il inline
#define fi first
#define se second
#define mk(a,b) make_pair(a,b)
#define numb (ch^'0')
#define pb push_back
#define solid const auto &
#define enter cout<<endl
#define pii pair<int,int>
using namespace std;
typedef long long ll;
template<class T>il void rd(T &x){
    char ch;x=0;bool fl=false;while(!isdigit(ch=getchar()))(ch=='-')&&(fl=true);
    for(x=numb;isdigit(ch=getchar());x=x*10+numb);(fl==true)&&(x=-x);}
template<class T>il void output(T x){if(x/10)output(x/10);putchar(x%10+'0');}
template<class T>il void ot(T x){if(x<0) putchar('-'),x=-x;output(x);putchar(' ');}
template<class T>il void prt(T a[],int st,int nd){for(reg i=st;i<=nd;++i) ot(a[i]);putchar('\n');}
namespace Modulo{
const int mod=1e9+7;
int ad(int x,int y){return (x+y)>=mod?x+y-mod:x+y;}
void inc(int &x,int y){x=ad(x,y);}
int mul(int x,int y){return (ll)x*y%mod;}
void inc2(int &x,int y){x=mul(x,y);}
int qm(int x,int y=mod-2){int ret=1;while(y){if(y&1) ret=mul(x,ret);x=mul(x,x);y>>=1;}return ret;}
}
using namespace Modulo;
namespace Miracle{
const int N=2003;
int A,B;
int f[N][N];
int C[N][N];
int main(){
    rd(A);rd(B);
    C[0][0]=1;
    int lim=max(A,B)+1;
    for(reg i=1;i<=lim;++i){
        C[i][0]=1;
        for(reg j=1;j<=i;++j){
            C[i][j]=ad(C[i-1][j],C[i-1][j-1]);
        }
    }
    for(reg i=1;i<=lim;++i){
        for(reg s=1;s<=lim;++s){
            f[i][s]=ad(f[i][s-1],C[i-1][s-1]);
        }
        for(reg s=1;s<=lim;++s) f[i][s]=ad(f[i][s],f[i][s-1]);
    }
    int ans=0;
    for(reg t=1;t<=A+1;++t){
        for(reg i=0;i<=t-1;++i){
            if(i!=0) ans=ad(ans,mul(C[B-1][i],f[i][t-i]));
            else ans=ad(ans,1);
        }
    }
    ot(ans);
    return 0;
}

}
signed main(){
    Miracle::main();
    return 0;
}

/*
   Author: *Miracle*
*/