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;
long long cnt;


typedef long long           ll;
typedef unsigned long long  ull;

#define sf                  scanf
#define pf                  printf
#define si(n)               scanf("%d",&n)
#define sii(x,y)            scanf("%d %d",&x,&y)
#define siii(x,y,z)         scanf("%d %d %d",&x,&y,&z)
#define sl(n)               scanf("%lld",&n)
#define sll(x,y)            scanf("%lld %lld",&x,&y)
#define slll(x,y,z)         scanf("%lld %lld %lld",&x,&y,&z)
#define yes                 printf("YES\n")
#define no                 printf("NO\n")
#define FOR(i,x,y)          for(int i=x;i<y;i++)
#define RFOR(i,x,y)         for(int i=x;i>=y;i--)
#define MAXN                100005
#define MOD                 1000000007
#define PI                  acos(-1)


int main ()
{

    ll tc;

    ll a,ans=1;

    cin>>tc;
    while(tc--)
    {
        cin>>a;
        ans*=a;


    }

    if(ans>1000000000000000000)
    {
        cout<<"-1"<<endl;
    }
    else
    {
        cout<<ans<<endl;
    }


    return 0;
}