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<cstdio>
#include<algorithm>
#include<queue>
#include<cmath>
#include<iostream>
#include<cstring>
#define ll long long
#define maxn 5005
			 
#define p 1000000007
#define re(i,a,b) for(int i=a;i<=b;i++)
#define for_edge(x) for(int i=head[x];i;i=e[i].nxt)
using namespace std;
inline int read(){char c=getchar();int f=1;int ans = 0;while(c>'9'||c<'0'){if(c=='-')f=-f;c=getchar();}while(c<='9'&&c>='0'){ans =ans*10+c-'0';c=getchar();}return ans*f;}
//_____________________________________________________________________________________________________
ll _C[maxn][maxn];
inline ll C(int n,int m)
{
	return (n>=0 && m>=0) ? _C[n][m] : 0;
}
int main()
{
	int A = read() , B = read();;
	_C[0][0] = 1;
	re(i,0,5000)
		re(j,0,i)
		{
			_C[i][j] %= p;
			_C[i+1][j] += _C[i][j];
			_C[i+1][j+1] += _C[i][j];
		}
	ll ans = 0;
	re(i,1,B)
	{
		re(j,1,B-i)
				( ans += C(B-1,i-1)*C(B-i-1,j-1)%p * C( A-2*B+2*i+j+2,2 ) )%=p;
		//( ans += C(B-1,i-1) *  )
	}
	ans += A+1;
	printf("%lld\n",ans);
    return 0;
}

