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>
#define sqr(x) ((x)*(x))
#define ll long long
#define ull unsigned long long
#define ui unsigned int
#define ld long double
#define vi vector<int>
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > > 
#define pii pair<int,int> 
#define pb push_back
#define mp make_pair
#define debuge cerr<<"isok"<<endl
#define debug(x) cerr<<#x<<"="<<x<<endl
#define dprintf(...) fprintf(stderr,__VA_ARGS__)
#define SS second
#define FF first
#define ls (k<<1)
#define rs (k<<1|1)
#define clr(a,x) memset(a,x,sizeof(a))
#define cpy(a,x) memcpy(a,x,sizeof(a))
#define file(x) freopen(x".in","r",stdin),freopen(x".out","w",stdout)
#define SZ(x) ((int)x.size())
using namespace std;
template<class T> inline void gmin(T &x,const T &y){if(x>y) x=y;}
template<class T> inline void gmax(T &x,const T &y){if(x<y) x=y;}
const int BufferSize=1<<16;
char buffer[BufferSize],*Bufferhead,*Buffertail;
bool Terminal;
inline char Getchar(){
	if(Bufferhead==Buffertail){
		int l=fread(buffer,1,BufferSize,stdin);
		if(!l){Terminal=1;return 0;}
		Buffertail=(Bufferhead=buffer)+l;
	}
	return *Bufferhead++;
}
template<class T>inline bool read(T &x){
	x=0;char c=Getchar(),rev=0;
	while(c<'0'||c>'9'){rev|=c=='-';c=Getchar();if(Terminal)return 0;}
	while(c>='0'&&c<='9') x=x*10+c-'0',c=Getchar();
	if(c=='.'){
		c=Getchar();double t=0.1;
		while(c>='0'&&c<='9') x=x+(c-'0')*t,c=Getchar(),t=t/10;
	}
	x=rev?-x:x;
	return 1;
}
template<class T1,class T2> inline bool read(T1 &x,T2 &y){return read(x)&read(y);}
template<class T1,class T2,class T3> inline bool read(T1 &x,T2 &y,T3 &z){return read(x)&read(y)&read(z);}
template<class T1,class T2,class T3,class T4> inline bool read(T1 &x,T2 &y,T3 &z,T4 &w){return read(x)&read(y)&read(z)&read(w);}
inline bool reads(char *x){
	char c=Getchar();
	while(c<33||c>126){c=Getchar();if(Terminal)return 0;}
	while(c>=33&&c<=126) (*x++)=c,c=Getchar();
	*x=0;return 1;
}
template<class T>inline void print(T x,const char c='\n'){
	if(!x){putchar('0');putchar(c);return;}
	if(x<0) putchar('-'),x=-x;
	int m=0,a[20];
	while(x) a[m++]=x%10,x/=10;
	while(m--) putchar(a[m]+'0');
	putchar(c);
}
//--------------------------------head---------------------------------------------

const int inf=0x3f3f3f3f;
const int N=4005,M=100005,mod=1e9+7;
template<class T,class S> inline void ch(T &x,const S y){x=(x+y)%mod;}
inline int exp(int x,int y,const int mod=::mod){
	int ans=1;
	while(y){
		if(y&1) ans=(ll)ans*x%mod;
		x=(ll)x*x%mod;y>>=1;
	}return ans;
}

int a,b,ans,c[N][N],s[N][N],f[N][N];

int main(){
#ifdef rqgao2014
	freopen("input.txt","r",stdin);
#endif
	scanf("%d%d",&a,&b);
	if(!a||!b){puts("1");return 0;}
	for(int i=0;i<=4000;i++){
		c[i][0]=c[i][i]=1;
		for(int j=1;j<i;j++)
			c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
		for(int j=i;~j;j--)
			s[i][j]=(s[i][j+1]+c[i][j])%mod;
	}
	for(int i=0;i<=a;i++)
		for(int j=0;j<=b;j++){
			if(!i){f[i][j]=1;continue;}
			f[i][j]=f[i-1][j];
			ch(f[i][j],s[j-1][max(j-i-1,0)]);
		}
//	printf("%d\n",f[a][b]);
	for(int i=0;i<=a;i++){
		int ta=a-i,tb=b-1;
		for(int j=max(tb-ta,0);j<=tb;j++)
			ch(ans,(ll)f[ta-tb+j][tb-j]*c[tb][j]);
	}
	printf("%d\n",ans);
	return 0;
}