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 <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <bitset>
#include <utility>
#include <assert.h>
using namespace std;
#define rank rankk
#define mp make_pair
#define pb push_back
#define xo(a,b) ((b)&1?(a):0)
#define tm tmp
//#pragma comment(linker, "/STACK:1024000000,1024000000") 
//#define LL ll
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef long long ll;
typedef pair<ll,int> pli;
typedef pair<ll,ll> pll;
const int INF=0x3f3f3f3f;
const ll INFF=0x3f3f3f3f3f3f3f3fll;
const int MAX=5e5+10;
//const ll MAXN=2e8;
//const int MAX_N=MAX;
const ll MOD=1e9+7;
//const long double pi=acos(-1.0);
//const double eps=0.00000001;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
template<typename T>inline T abs(T a) {return a>0?a:-a;}
template<class T> inline
void read(T& num) {
    bool start=false,neg=false;
    char c;
    num=0;
    while((c=getchar())!=EOF) {
        if(c=='-') start=neg=true;
        else if(c>='0' && c<='9') {
            start=true;
            num=num*10+c-'0';
        } else if(start) break;
    }
    if(neg) num=-num;
}
inline ll powMM(ll a,ll b,ll M){
    ll ret=1;
    a%=M;
//    b%=M;
    while (b){
        if (b&1) ret=ret*a%M;
        b>>=1;
        a=a*a%M;
    }
    return ret;
}
void open()
{
//    freopen("1009.in","r",stdin);
    freopen("out.txt","w",stdout);
}

const int N = 2020;
ll C[N][N];
void init() {
        for (int i = 0; i < N; i ++) C[i][0] = C[i][i] = 1;
        for (int i = 1; i < N; i ++) {
                for (int j = 1; j < i; j ++) {
                        C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
                }
        }
}
int a,b;
ll an,lin;
int main()
{
    init();
    scanf("%d%d",&a,&b);
    an=a+1;
    for(int i=1;i<=min(a,b-1);i++)//第一次向下的位置的横坐标
    {
        for(int j=0;j<i&&j+i<=a;j++)
        {
            lin=C[b-1][i]*C[i-1][j]%MOD*C[a-i-j+2][2]%MOD;
            an=(an+lin)%MOD;
        }
    }
    printf("%lld\n",an);

}
/*
8
11011011
*/
