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 Pascal (FPC 3.0.4) (judge verdict: Runtime Error)

program hi;
var 
    ans: QWord;
    a: array[1..10010] of QWord;
    n, i: integer;

procedure mergesort(l, r : integer);
var 
    tmp : array [1..100] of QWord;
    mid, p1, p2, pt, i : integer;
begin
    if l < r then begin
        mid := (l + r) div 2;
        mergesort(l, mid);
        mergesort(mid + 1, r);
        p1 := l; p2 := mid + 1; pt := l;
        while (p1 <= mid) and (p2 <= r) do begin
            if a[p1] < a[p2] then begin
                tmp[pt] := a[p1];
                inc(p1);
            end else begin
                tmp[pt] := a[p2];
                inc(p2);
            end;
            inc(pt);
        end;
        while p1 <= mid do begin
            tmp[pt] := a[p1];
            inc(pt); inc(p1);
        end;
        while p2 <= r do begin
            tmp[pt] := a[p2];
            inc(pt); inc(p2);
        end;
        for i := l to r do 
            a[i] := tmp[i];
    end;
end;

begin
    readln(n);
    ans := 1;
    for i := 1 to n do 
        read(a[i]);
    mergesort(1, n);
    for i := 1 to n do begin
        if a[i] = 0 then begin
            writeln(0);
            exit;
        end;
        if ans <= 1000000000000000000 div a[i] then 
            ans := ans * a[i]
        else begin
            writeln(-1);
            exit;
        end;
    end;
    writeln(ans);
end.