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;
#define IOS ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep1(i,n) for(ll i=1;i<=n;i++)
#define rep_start(a,i,n) for(ll i=a;i<=n;i++)
#define rev(i,n) for(ll i=n-1;i>=0;i--)
#define rev1(i,n) for(ll i=n;i>0;i--)
#define sz(a) int((a).size())
#define all(c) c.begin(),c.end()
#define tr(c,i) for(typeof(c).begin() i+=c.begin; i!=c.end(); i++)
#define present(c,x) (c.find(x)!=c.end())
#define cpresent(c,x) (find(all(c),x)!=c.end())
#define pb push_back
#define F first
#define S second
#define INF INT_MAX
#define mod 1000000007
#define print1(a) cout<<a<<"\n";
#define print2(a,b) cout<<a<<" "<<b<<"\n";
#define print3(a,b,c) cout<<a<<" "<<b<<" "<<c<<"\n";
#define mt make_tuple
#define eb emplace_back

typedef long long int ll;
typedef vector<ll> vi;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<vi> vvi;
typedef pair<ll,ll> ii;
typedef vector<ii> vii;
typedef tuple<int,int,int> State;

void solve(){
    ll n,ans=1;cin>>n;
    vi v(n);
    bool c=0;
    rep(i,n) {cin>>v[i];if(v[i]==0) c=1;}
    if(c) {cout<<0;return;}
    rep(i,n) {
        ans*=v[i];
        if(ans>(ll)1e18) {cout<<-1;return;}
    }
    cout<<ans;
}

signed main() {
    // your code goes here
    IOS
    ll t=1;
    //cin>>t;
    rep1(i,t){
        solve(); 
    }
    return 0;
}