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:
#pragma region
#ifndef M207
#pragma GCC optimize("O2,Ofast,inline,unroll-all-loops,-ffast-math")
#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#endif
#include <bits/stdc++.h>
using namespace std;
#define ri register int
#define il inline
#define LL long long
#define uint unsigned int
#define ull unsigned long long
#define solid const auto &
#define pb emplace_back
#define mp make_pair
#define pii pair<int,int>
#define pll pair<LL,LL>
#define fi first
#define se second
#define gm int mid((l+r)/2)
#define space putchar(' ')
#define enter putchar('\n')
#define rd() ({ri t;in(t);t;})
#define Size(x) ((int)x.size())
#define mem(x,y) memset(x,y,sizeof(x))
template<class T>il void in(T &x)
{
	x=0; char c=getchar(); bool f=0;
	while(!isdigit(c)) f|=(c=='-'),c=getchar();
	while(isdigit(c)) x=x*10+(c^'0'),c=getchar();
	f?x=-x:0;
}
template<class T>il void out(T x,const char c='\n')
{
	static short st[30]; short m=0;
	if(x<0) putchar('-'),x=-x;
	do st[++m]=x%10,x/=10; while(x);
	while(m) putchar(st[m--]|'0');
	putchar(c);
}
template<class T>il void err(const T &x,const char c='\n') {cerr<<x<<c;}
template<class T,class ...Args>il void in(T &x,Args &...args) {in(x); in(args...);}
template<class T,class ...Args>il void out(const T &x,const Args &...args) {out(x,' '); out(args...);}
template<class T,class ...Args>il void err(const T &x,const Args &...args) {err(x,' '); err(args...);}
template<class T>il void prt(T a[],int n) {for(ri i=0; i<n; ++i) out(a[i],i==n-1?'\n':' ');}
template<class T>il void clr(T a[],int n) {memset(a,0,sizeof(T)*n);}
template<class T>il void clr(T *a,T *b) {memset(a,0,sizeof(T)*(b-a));}
template<class T>il bool ckmax(T &a,const T &b) {return a<b?a=b,1:0;}
template<class T>il bool ckmin(T &a,const T &b) {return a>b?a=b,1:0;}
namespace MOD_CALC
{
const int md=1e9+7,inv2=(md+1)/2;
il int add(const int a,const int b) {return a+b>=md?a+b-md:a+b;}
il int sub(const int a,const int b) {return a-b<0?a-b+md:a-b;}
il int mul(const int a,const int b) {return (LL)a*b%md;}
il void inc(int &a,const int b) {(a+=b)>=md?a-=md:0;}
il void dec(int &a,const int b) {(a-=b)<0?a+=md:0;}
il int qpow(int a,int b) {int r=1; for(; b; b>>=1,a=mul(a,a)) if(b&1) r=mul(r,a); return r;}
il int qpow(int a,int b,const int p) {int r=1; for(; b; b>>=1,a=(LL)a*a%p) if(b&1) r=(LL)r*a%p; return r;}
il int mdinv(const int a) {return qpow(a,md-2);}
template<class ...Args>il int add(const int a,const int b,const Args &...args) {return add(add(a,b),args...);}
template<class ...Args>il int mul(const int a,const int b,const Args &...args) {return mul(mul(a,b),args...);}
} using namespace MOD_CALC;
namespace i207M
{
#pragma endregion
#define N 4005
int c[N][N],sc[N][N];
void init(int n)
{
	c[0][0]=sc[0][0]=1;
	for(ri i=1; i<=n; ++i)
	{
		c[i][0]=1;
		for(ri j=1; j<=i; ++j)
			c[i][j]=add(c[i-1][j-1],c[i-1][j]);
		for(ri j=i; j>=0; --j)
			sc[i][j]=add(sc[i][j+1],c[i][j]);
	}
}
int A,B;
signed main()
{
#ifdef M207
	freopen("in.in","r",stdin);
	// freopen("ot.out","w",stdout);
#endif
	in(A,B);
	if(B==1)
	{
		out(A+1);
		return 0;
	}
	int len=A+B,ans=0; init(len);
	for(ri q=B+1; q<=len; ++q) // q = [p+B,len]
	{
		// max(1,q+B-len-1),B-1
		for(ri i=max(1,q+B-len-1); i<B; ++i)
		{
			int t=2*B-2*i-len+q-2;
			inc(ans,mul(c[B-1][i-1],sc[B-i-1][max(t,0)],min(A+1,q-B)));
			// for(ri q=p+B; q+B-i-1<=len; ++q)
			// 	for(ri j=1; j<=B-i; ++j)
			// 	{
			// 		if(!(B-i-j<=len-(q+B-i-1))) continue;
			// 		inc(ans,mul(C(B-1,i-1),C(B-i-1,j-1)));
			// 	}
		}
	}
	inc(ans,A+1);
	out(ans);
	return 0;
}
#pragma region
}
signed main()
{
	i207M::main();
	return 0;
}
#pragma endregion