Problem:
Score : 100 points

Problem Statement

Having learned the multiplication table, Takahashi can multiply two integers between 1 and 9 (inclusive) together. He cannot do any other calculation.

Given are two integers A and B.

If Takahashi can calculate A \times B, print the result; if he cannot, print -1 instead.

Constraints

1 \leq A \leq 20

1 \leq B \leq 20

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

If Takahashi can calculate A \times B, print the result; if he cannot, print -1.

Sample Input 1

2 5

Sample Output 1

10

2 \times 5 = 10.

Sample Input 2

5 10

Sample Output 2

-1

5\times 10 = 50, but Takahashi cannot do this calculation, so print -1 instead.

Sample Input 3

9 9

Sample Output 3

81

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

#include <bits/stdc++.h>
#ifdef LOCAL_DEBUG
#define DEBUG 1
#define CERR if(DEBUG) cerr
#define MACRO(_1,_2,_3,_4,_5,_6,_7,_8,_9,_10,NAME,...) NAME
#define pr(...) CERR<<MACRO(__VA_ARGS__,pr10,pr9,pr8,pr7,pr6,pr5,pr4,pr3,pr2,pr1)(__VA_ARGS__)<<endl
#define pr1(a) (#a)<<"="<<(a)<<" "
#define pr2(a,b) pr1(a)<<pr1(b)
#define pr3(a,b,c) pr1(a)<<pr2(b,c)
#define pr4(a,b,c,d) pr1(a)<<pr3(b,c,d)
#define pr5(a,b,c,d,e) pr1(a)<<pr4(b,c,d,e)
#define pr6(a,b,c,d,e,f) pr1(a)<<pr5(b,c,d,e,f)
#define pr7(a,b,c,d,e,f,g) pr1(a)<<pr6(b,c,d,e,f,g)
#define pr8(a,b,c,d,e,f,g,h) pr1(a)<<pr7(b,c,d,e,f,g,h)
#define pr9(a,b,c,d,e,f,g,h,i) pr1(a)<<pr8(b,c,d,e,f,g,h,i)
#define pr10(a,b,c,d,e,f,g,h,i,j) pr1(a)<<pr9(b,c,d,e,f,g,h,i,j)
#define prArr(a){CERR<<(#a)<<"={";int i=0;for(auto t:(a))CERR<<(i++?", ":"")<<t;CERR<<"}"<<endl;}
#else
#define DEBUG 0
#define pr(...)
#define prArr(a)
#endif
using namespace std;
using Int = long long;
using _int = int;
using ll = long long;
using Double = long double;
const Int INF = (1LL<<60)+1e9; // ~ 1.15 * 1e18
const Int mod = (1e9)+7;
const Double EPS = 1e-8;
const Double PI = 6.0 * asin((Double)0.5);
using P = pair<Int,Int>;
template<class T> T Max(T &a,T b){return a=max(a,b);}
template<class T> T Min(T &a,T b){return a=min(a,b);}
template<class T1, class T2> ostream& operator<<(ostream& o,pair<T1,T2> p){return o<<"("<<p.first<<","<<p.second<<")";}
template<class T1, class T2, class T3> ostream& operator<<(ostream& o,tuple<T1,T2,T3> t){return o<<"("<<get<0>(t)<<","<<get<1>(t)<<","<<get<2>(t)<<")";}
template<class T1, class T2> istream& operator>>(istream& i,pair<T1,T2> &p){return i>>p.first>>p.second;}
template<class T> ostream& operator<<(ostream& o,vector<T> a){Int i=0;for(T t:a)o<<(i++?" ":"")<<t;return o;}
template<class T> istream& operator>>(istream& i,vector<T> &a){for(T &t:a)i>>t;return i;}
//INSERT ABOVE HERE


signed main(){
  srand((unsigned)time(NULL));
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(12);

  Int A, B;
  cin>>A>>B;

  if(1 <= A && A <= 9 && 1 <= B && B <= 9) cout<<A * B<<endl;
  else cout<<-1<<endl;

  return 0;
}