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)

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

#define CPP_STR(x) CPP_STR_I(x)
#define CPP_CAT(x, y) CPP_CAT_I(x, y)
#define CPP_STR_I(args...) #args
#define CPP_CAT_I(x, y) x##y

#define ASSERT(expr...) assert((expr))

using i8 = int8_t;
using u8 = uint8_t;
using i16 = int16_t;
using u16 = uint16_t;
using i32 = int32_t;
using u32 = uint32_t;
using i64 = int64_t;
using u64 = uint64_t;

using f32 = float;
using f64 = double;
// }}}

constexpr i64 INF = 1'010'000'000'000'000'017LL;

constexpr i64 MOD = 1'000'000'007LL;

constexpr f64 EPS = 1e-12;

constexpr f64 PI = 3.14159265358979323846;

#define M5 100007
#define M9 1000000000

// util {{{
#define FOR(i, start, end) for (i64 i = (start), CPP_CAT(i, xxxx_end) = (end); i < CPP_CAT(i, xxxx_end); ++i)
#define REP(i, n) FOR(i, 0, n)

#define ALL(f, c, ...) (([&](decltype((c)) cccc) { return (f)(std::begin(cccc), std::end(cccc), ##__VA_ARGS__); })(c))
#define ll long long int
#define VI vector<int>
template <typename C>
i64 SIZE(const C &c)
{
    return static_cast<i64>(c.size());
}

template <typename T, size_t N>
i64 SIZE(const T (&)[N]) { return static_cast<i64>(N); }

template <typename T, typename U, typename Comp = less<>>
bool chmax(T &xmax, const U &x, Comp comp = {})
{
    if (comp(xmax, x))
    {
        xmax = x;
        return true;
    }
    return false;
}

template <typename T, typename U, typename Comp = less<>>
bool chmin(T &xmin, const U &x, Comp comp = {})
{
    if (comp(x, xmin))
    {
        xmin = x;
        return true;
    }
    return false;
}
// }}}

// init {{{
struct ProconInit
{
    static constexpr int IOS_PREC = 15;
    static constexpr bool AUTOFLUSH = false;

    ProconInit()
    {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(IOS_PREC);
        if (AUTOFLUSH)
            cout << unitbuf;
    }
} PROCON_INIT;
// }}}

//--------------------------------------------------------------------

bool isOK(int mid, double key)
{
}

int binary_search(double key)
{
    int left = -1; //「index = 0」が条件を満たすこともあるので、初期値は -1
    int right = 2000000000;
    /* どんな二分探索でもここの書き方を変えずにできる！ */
    while (right - left > 1)
    {
        int mid = left + (right - left) / 2;

        if (isOK(mid, key))
            right = mid;
        else
            left = mid;
    }

    /* left は条件を満たさない最大の値、right は条件を満たす最小の値になっている */
    return right;
}

bool isKaibun(string st)
{
    bool ans = true;
    int n = st.length();
    REP(i, n / 2)
    {
        if (st[i] != st[n - i - 1])
        {
            return false;
        }
    }
    return true;
}

int toInt(char a)
{
    return a - '0';
}

int gcd(int a, int b)
{
    if (a % b == 0)
    {
        return (b);
    }
    else
    {
        return (gcd(b, a % b));
    }
}

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

void solve()
{
    int N;
    cin >> N;
    vector<double> A(N);
    REP(i, N)
    {
        cin >> A[i];
        if (A[i] == 0)
        {
            cout << 0;
            return;
        }
    }

    sort(A.begin(), A.end(), greater<>());
    long long ans = A[0];
    FOR(i, 1, N)
    {
        ans *= (long long)A[i];
        //        cout << ans << endl;
        if (ans > 1000000000000000000 || ans < 0)
        {
            cout << -1;
            return;
        }
    }
    cout << fixed << setprecision(0) << ans;
}

signed main()
{

    solve();

    return 0;
}