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)

// Template //
#include<bits/stdc++.h>
using namespace std;

// マクロ //
#define rep(i,N) for(int i = 0; i < N; i++)
#define all(x) x.begin(), x.end()
#define sort(x) sort(all(x))
#define cou(x) cout << x << endl
#define y() cout << "Yes" << endl
#define n() cout << "No" << endl
#define Y() cout << "YES" << endl
#define N() cout << "NO" << endl
#define x2(x) (x) * (x)

// 型エイリアス //
using lint = long long;
using vi = vector<int>;
using vli = vector<lint>;
using vc = vector<char>;
using vs = vector<string>;
using vb = vector<bool>;
using vvi = vector<vi>;
using pii = pair<int, int>;
using vpii = vector<pii>;
using msi = map<string, int>;

// 関数 //
int gcd(int a, int b) {
    int t;
    while (b != 0) {
        t = a % b;
        a = b;
        b = t;
    }
    return a;
}

int lcm(int a, int b) {
    return a * b / gcd(a, b);
}

double distance(pii a, pii b) {
    double dist;
    dist = sqrt(x2(a.first - b.first) + x2(a.second - b.second));
    return dist;
}

lint perm(int a) {
    lint perm = 1;
    for (int i = a; i >= 1; i--) {
        perm *= i;
    }
    return perm;
}

lint comb(int n, int m) {
    return perm(n) / (perm(n - m) * perm(m));
}

// End of Template //



int main() {

    int N , M;
    cin >> N >> M;

    cou((N < 10 && M < 10 ? N * M : -1));
}