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>
using namespace std;
namespace mine
{
    typedef long long ll;
    #define pr pair<int,int>
    #define FR first
    #define SE second
    #define MP make_pair
    #define PB push_back
    #define vc vector
    #define all(x) (x).begin(),(x).end()
    #define sz(x) ((int)(x).size())
    #define bin(x) (1ll<<(x))
    #define GG(x) if(x) puts("error")
    ll qread()
    {
        ll ans=0,f=1;char c=getchar();
        while(c<'0' or c>'9') {if(c=='-')f=-1;c=getchar();}
        while('0'<=c and c<='9') ans=ans*10+c-'0',c=getchar();
        return ans*f;
    }
    void write(ll num)
    {
        if(num<0) putchar('-'),num=-num;
        if(num>=10) write(num/10);
        putchar('0'+num%10);
    }
    void write1(ll num){write(num);putchar(' ');}
    void write2(ll num){write(num);putchar('\n');}
    template<typename T> void chmax(T &x,const T y) {x=(x>y?x:y);}
    template<typename T> void chmin(T &x,const T y) {x=(x<y?x:y);}
    ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}

    const int INF=0x3f3f3f3f;
    const int MOD=1e9+7;
    inline int mm(const int x){return x>=MOD?x-MOD:x;}
    template<typename T> inline void add(T &x,const T y){x=mm(x+y);}
    inline ll qpower(ll x,ll e)
    {
        ll ans=1;GG(e<0);
        while(e){if(e&1)ans=ans*x%MOD;x=x*x%MOD;e>>=1;}return ans;
    }
    ll invm(ll x){return qpower(x,MOD-2);}
    const int N=2e3+10;

    int C[N][N],pp[N][N],solve1[N][N];
    void main()
    {
        C[0][0]=1;for(int i=0;i<N;i++) pp[0][i]=solve1[0][i]=1;
        for(int i=1;i<N;i++)
        {
            C[i][0]=pp[i][0]=solve1[i][0]=1;
            for(int j=1;j<N;j++)
            {
                C[i][j]=mm(C[i-1][j-1]+C[i-1][j]);
                pp[i][j]=mm(pp[i][j-1]+C[i][j]);
            }
        }
        for(int i=1;i<N;i++) for(int j=1;j<N;j++) solve1[i][j]=mm(solve1[i-1][j]+pp[j-1][i]);

        int A=qread(),B=qread();ll ans=0;
        for(int t=1;t<=A+1;t++) for(int i=0;i<=t-1;i++) add(ans, 1ll*C[B-1][i]*solve1[t-1-i][i]%MOD );write(ans);
    }//(ans+MOD)%MOD
};
int main()
{
    srand(time(0));
    mine::main();
}