Problem:
Score : 100 points

Problem Statement

We have two integers: A and B.

Print the largest number among A + B, A - B, and A \times B.

Constraints

All values in input are integers.

-100 \leq A,\ B \leq 100

Input

Input is given from Standard Input in the following format:

A B

Output

Print the largest number among A + B, A - B, and A \times B.

Sample Input 1

-13 3

Sample Output 1

-10

The largest number among A + B = -10, A - B = -16, and A \times B = -39 is -10.

Sample Input 2

1 -33

Sample Output 2

34

The largest number among A + B = -32, A - B = 34, and A \times B = -33 is 34.

Sample Input 3

13 3

Sample Output 3

39

The largest number among A + B = 16, A - B = 10, and A \times B = 39 is 39.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _GLIBCXX_DEBUG
#include<cmath>
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
vector<int> arr;
stack<int> st;
queue<int> qu;
queue<pair<int,int> > qu2;
priority_queue<int> pq;
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define rep2(i,n) for(int i=1; i<=(int)(n); i++)
typedef set<int> set_t;
typedef set<string> set_g;
typedef complex<double> xy_t;
static const int NIL = -1;

/*int P[10010];
void init(int n){
    for(int i=0;i<n;++i)P[i]=i;
}
int root(int a) {
    if(P[a]==a)return a;
    return (P[a]=root(P[a]));
}
bool is_same_set(int a,int b){
    return root(a) == root(b);//代表元を求める
}
void unite(int a,int b){
    P[root(a)]=root(b);
}//対set,グラフ？
//セグメント木？？
*/

double dot_product(xy_t a, xy_t b) { return (conj(a)*b).real(); }//内積
double cross_product(xy_t a, xy_t b) { return (conj(a)*b).imag(); }//外積
xy_t projection(xy_t p, xy_t b) { return b*dot_product(p,b)/norm(b); }//投影
//対図形

#define mod 1000000007
/*ll f[2001];
//int n,k とかしておく

ll pw(ll x, ll y){//euclidの互除法より
    ll a= 1;
    while(y){
        if(y&1){//奇数なら
            a = a*x%mod;
        }
        x = x*x%mod;
        y /= 2;
    }
    return a;
}

ll modinv(ll x){//逆元を求める
    return pw(x, mod - 2 );
}

ll comb(int n,int r){
    if(n<r){
        return 0;
    }
    return f[n] * modinv(f[r])%mod*modinv(f[n-r])%mod;
}//対combination//ただしfは用意してね
*/

struct BIT{
    vector<int> dat;
    int sz;
    BIT(int n){
        for(sz=1;sz<n;sz *= 2);
        dat.resize(++sz);
    }
    int q(int n){
        int ret = 0;
        for(int i=n;i>0;i-=i&-i)ret +=dat[i];//和の計算　iから最後の１のbit(i&-i 多分＆はビット積)を減算
        for(int i=n;i<sz;i+=i&-i) dat[i]++;//値の加算　iから最後mの１のbitを加算
        return ret;
    }
};

int main(){
    int a,b;
    cin >> a>> b;
    cout<<max(a+b,max(a-b,a*b))<<endl;
    return 0;
}