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: Accepted)

#include <iostream>
#include <vector>
#include <string>
#include <numeric>
#include <tuple>
#include <algorithm>
#include <array>
#include <cmath>
using namespace std;
using ll = long long;
using ull = unsigned long long;
#define rep(i,n) for(decltype(n) i=0; i<n; ++i)
inline constexpr ll Inf = (1ULL << 62) -1;

template <class T>
void updatemax(T& a, T b) { if (b > a) a = b; }

int main() {
    ull ans=1;
    ll N;
    cin >> N;
    vector<ull> a(N);
    rep(i,N) cin >> a[i];
    if (find(begin(a), end(a), 0) != end(a)) {
        ans = 0;
    } else {
        double log10 = log(10);
        rep(i,N) {
            ull p = ans;
            ans *= a[i];
            if ((log(p)+log(a[i]))/log10 > 18.000001 || ans > 1000000000000000000 || ans < p) {
                cout << -1 << endl;
                return 0;
            }
        }
    }
    cout << ans << endl;
    return 0;
}