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>
using namespace std;
typedef long long ll;
inline void readInt(int &x) {
	x=0;int f=1;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	x*=f;
}
inline void readLong(ll &x) {
	x=0;int f=1;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	x*=f;
}
/*================Header Template==============*/
const ll mod=1e9+7;
const int maxn=4005;
ll fac[maxn][maxn],a,b,ans=0;
inline void init() {
	memset(fac,0,sizeof(fac));
	fac[0][0]=1;
	for(int i=1;i<=4000;i++) {
		fac[i][0]=1;
		for(int j=1;j<=i;j++)
			fac[i][j]=(fac[i-1][j-1]+fac[i-1][j])%mod;
	}
}
inline bool check(int x,int y){return (b-x)+(b-x-y)<=a;}
inline int getindex(int x,int y){return a-(b-x)-(b-x-y)+2;}
inline void solve() {
	for(int cnta=1;cnta<b;cnta++)
		for(int cntb=1;cntb<=b-cnta;cntb++)
			if(check(cnta,cntb)) {
				ll tmp=fac[b-1][cnta-1]*fac[b-cnta-1][cntb-1]%mod;
				tmp=tmp*fac[getindex(cnta,cntb)][2]%mod;
				ans=(ans+tmp)%mod;
			}
	ans=(ans+a+1)%mod;
}
int main() {
	init();
	readLong(a);
	readLong(b); 
	solve();
	printf("%lld\n",ans);
	return 0;
}
