#include<bits/stdc++.h>

using namespace std;

#define LL long long
#define DD double
#define Pb push_back
#define Bp pop_back
#define Pf push_front
#define Fp pop_front
#define Ub upper_bound
#define Lb lower_bound
#define In insert
#define All(x) x.begin(), x.end()
#define mem(a, b) memset(a, b, sizeof(a))
#define fast ios_base::sync_with_stdio(0);cin.tie(0)
#define X first
#define Y second
#define mx1 10
#define mx2 110
#define mx3 1010
#define mx4 10010
#define mx5 100010
#define mx6 1000010

typedef vector<LL> Vl;
typedef vector<DD> Vd;
typedef vector<bool> Vb;
typedef vector<Vl> VVl;
typedef pair<LL, LL> Pll;
typedef pair<DD, DD> Pdd;
typedef vector<Pll> Vpl;
typedef vector<Pdd> Vpd;
typedef queue<LL> Ql;
typedef stack<LL> Sl;
typedef deque<LL> Dl;

LL _set(LL N, LL pos) {return N = N | (1 << pos);}
LL _reset(LL N, LL pos) {return N = N & ~(1 << pos);}
bool _check(LL N, LL pos) {return (bool)(N & (1 << pos));}
bool _upper(char a) {return a >= 'A' && a <= 'Z';}
bool _lower(char a) {return a >= 'a' && a <= 'z';}
bool _digit(char a) {return a >= '0' && a <= '9';}

LL dx[] = {1, -1, 0, 0, -1, -1, 1, 1};
LL dy[] = {0, 0, 1, -1, -1, 1, -1, 1};

///**************************DONE****************************///

string multiply(string num1, string num2)
{
    int len1 = num1.size();
    int len2 = num2.size();
    if (len1 == 0 || len2 == 0)
    return "0";

    // will keep the result number in vector
    // in reverse order
    vector<int> result(len1 + len2, 0);

    // Below two indexes are used to find positions
    // in result.
    int i_n1 = 0;
    int i_n2 = 0;

    // Go from right to left in num1
    for (int i=len1-1; i>=0; i--)
    {
        int carry = 0;
        int n1 = num1[i] - '0';

        // To shift position to left after every
        // multiplication of a digit in num2
        i_n2 = 0;

        // Go from right to left in num2
        for (int j=len2-1; j>=0; j--)
        {
            // Take current digit of second number
            int n2 = num2[j] - '0';

            // Multiply with current digit of first number
            // and add result to previously stored result
            // at current position.
            int sum = n1*n2 + result[i_n1 + i_n2] + carry;

            // Carry for next iteration
            carry = sum/10;

            // Store result
            result[i_n1 + i_n2] = sum % 10;

            i_n2++;
        }

        // store carry in next cell
        if (carry > 0)
            result[i_n1 + i_n2] += carry;

        // To shift position to left after every
        // multiplication of a digit in num1.
        i_n1++;
    }

    // ignore '0's from the right
    int i = result.size() - 1;
    while (i>=0 && result[i] == 0)
    i--;

    // If all were '0's - means either both or
    // one of num1 or num2 were '0'
    if (i == -1)
    return "0";

    // generate the result string
    string s = "";

    while (i >= 0)
        s += std::to_string(result[i--]);

    return s;
}

int main()
{
    //fast;
    int n;
    cin >> n;

    string s = "1", t = "1000000000000000000", x[mx5];
    for(int i = 0; i < n; i++){
        cin >> x[i];

        if(x[i] == "0")
            return cout << 0 << '\n', 0;
    }

    for(int i = 0; i < n; i++){
        s = multiply(s, x[i]);
        if(s.size() > 19)
            return cout << -1 << '\n', 0;
    }

    if(s == t)
        cout << s << '\n';
    else if(s.size() > 18)
        cout << -1 << '\n';
    else
        cout << s << '\n';
    return 0;
}


Measured memory_bytes: 11340