Problem:
Score : 200 points

Problem Statement

Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.

However, if the result exceeds 10^{18}, print -1 instead.

Constraints

2 \leq N \leq 10^5

0 \leq A_i \leq 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

Print the value A_1 \times ... \times A_N as an integer, or -1 if the value exceeds 10^{18}.

Sample Input 1

2
1000000000 1000000000

Sample Output 1

1000000000000000000

We have 1000000000 \times 1000000000 = 1000000000000000000.

Sample Input 2

3
101 9901 999999000001

Sample Output 2

-1

We have 101 \times 9901 \times 999999000001 = 1000000000000000001, which exceeds 10^{18}, so we should print -1 instead.

Sample Input 3

31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0

Sample Output 3

0

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb push_back
#define pi (acos(-1))
#define ull  unsigned long long
#define ld long double
///freopen("input.txt","r",stdin);
///freopen("output.txt","w",stdout);
#define M 100000
ll LCM(ll a, ll b)
{
    ll g = __gcd(a,b);
    return (a*b)/g ;
}
string numtostr(ll n)
{
    ostringstream str1 ;
    str1 << n ;
    return str1.str();
}
ll strtonum(string s)
{
    ll x ;
    stringstream str1(s);
    str1 >> x ;
    return x ;
}
int n,i;
ull s=1;
bool b=true;
int main(void)
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cin>>n;
    ull a[n];
    for(i=0; i<n; ++i)
    {
        cin>>a[i];
        if(a[i]==0)
            b=false;
    }
    if(b==false)
        cout<<0<<"\n";
    else
    {
        for(i=0; i<n; ++i)
        {
            s*=a[i];
            if(s>1000000000000000000)
            {
                cout<<-1<<"\n";
                return 0;
            }
        }
        cout<<s<<"\n";
    }

    return 0;
}