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>
    typedef long long ll;
    using namespace std;
    inline ll read(){
    	ll f=1,x=0;char ch;
    	do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
    	do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
    	return f*x;
    }
    const int N=2010;
    const int yql=1e9+7;
    int s[N][N],f[N][N],c[N][N];
    inline void pre(){
    	for(int i=0;i<N;i++){
    		f[i][0]=1;s[i][0]=1;c[i][0]=1;
    		for(int j=1;j<i+1;j++)c[i][j]=(c[i-1][j]+c[i-1][j-1])%yql;
    		for(int j=1;j<N;j++)s[i][j]=(s[i][j-1]+c[i][j])%yql;
    		for(int j=1;j<N;j++)f[i][j]=(f[i][j-1]+s[i][j])%yql;
    	}
    }
    inline int solve(int t,int b){
    	int ans=1;
    	for(int i=1;i<=min(t-1,b);i++)ans=(ans+(1LL*f[i-1][t-i-1]*c[b][i])%yql)%yql;
    	return ans;
    }
    int main(){
    	//freopen("dick.in","r",stdin);
    	//freopen("dick.out","w",stdout);
    	int a=read(),b=read();
    	int ans=0;pre();
    	for(int i=1;i<=a+1;i++)ans=(ans+solve(i,b-1))%yql;
    	printf("%d\n",ans);
    	return 0;
    }