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 int ll;
typedef unsigned int uint;
typedef unsigned char uchar;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;

#define REP(i,x) for(int i=0;i<(int)(x);i++)
#define REPS(i,x) for(int i=1;i<=(int)(x);i++)
#define RREP(i,x) for(int i=((int)(x)-1);i>=0;i--)
#define RREPS(i,x) for(int i=((int)(x));i>0;i--)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();i++)
#define RFOR(i,c) for(__typeof((c).rbegin())i=(c).rbegin();i!=(c).rend();i++)
#define ALL(container) (container).begin(), (container).end()
#define RALL(container) (container).rbegin(), (container).rend()
#define SZ(container) ((int)container.size())
#define mp(a,b) make_pair(a, b)
#define pb push_back
#define eb emplace_back
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );
#define __builtin_popcount __builtin_popcountll

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
template<class T> ostream& operator<<(ostream &os, const vector<T> &t) {
    os<<"["; FOR(it,t) {if(it!=t.begin()) os<<","; os<<*it;} os<<"]"; return os;
}
template<class T> ostream& operator<<(ostream &os, const set<T> &t) {
    os<<"{"; FOR(it,t) {if(it!=t.begin()) os<<","; os<<*it;} os<<"}"; return os;
}
template<class S, class T> ostream& operator<<(ostream &os, const pair<S,T> &t) { return os<<"("<<t.first<<","<<t.second<<")";}
template<class S, class T> pair<S,T> operator+(const pair<S,T> &s, const pair<S,T> &t){ return pair<S,T>(s.first+t.first, s.second+t.second);}
template<class S, class T> pair<S,T> operator-(const pair<S,T> &s, const pair<S,T> &t){ return pair<S,T>(s.first-t.first, s.second-t.second);}

const int INF = 1<<28;
const double EPS = 1e-8;
const int MOD = 1000000007;

template<int N, int MOD>
struct Comb{
    ll fact[N+1], inv[N+1];
    Comb(){
        fact[0] = inv[0] = inv[1] = 1;
        for(int i=2;i<=N;i++) inv[i] = (MOD+((ll)-inv[MOD%i]*(MOD/i))%MOD)%MOD;
        REPS(i, N){
            fact[i] = fact[i-1]*i%MOD;
            inv[i] = inv[i]*inv[i-1]%MOD;
        }
    }
    ll operator()(int n, int r){
        if(n < 0 || r < 0 || r > n) return 0;
        ll res = inv[r]*inv[n-r];
        if(res >= MOD) res%= MOD;
        res *= fact[n];
        if(res >= MOD) res%= MOD;
        return res;
    }
};
Comb<10000, MOD> C;

int T, a, b;
ll table[2048][2048];

ll solve(int a, int b){
    if (b == 0) return 1;
    ll ans = 0;
    REP(i, a + 1) { // 最初からi個連続でr
        // red*s + blue*(b - 1)
        ans += table[b - 1][min(b, a - i + 1)];
    }
    return ans % MOD;
}


int main(int argc, char *argv[]){
    ios::sync_with_stdio(false);

    REP(i, 2011) {
        REP(j, 2010) {
            table[i][j + 1] = (table[i][j] + C(i, j)) % MOD;
        }
    }
    cin >> a >> b;
    ll ans = 0;
    REP(i, a + 1) { // 最初からi個連続でr
        // red*s + blue*(b - 1)
        int s = a - i;
        REP(j, min(b, s + 1)) {
            // b-1個取った中にrがj個含まれていた
            // red * (s - j) + blue * j
            ans = (ans + C(b - 1, j) * solve(s - j, j)) % MOD;
        }
    }
    cout << ans % MOD << endl;
    return 0;
}
