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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

#define SIZE 100005
#define PI acosl(-1) //3.14159265358979323846264338327950L 
#define rep(i, N) for (i = 0; i < N; i++) //制御変数iを用いてN回転
#define array(N, t) (t*)calloc(N, sizeof(t)) //t型N要素の1次元配列を動的確保後0クリア(freeを忘れずに)
#define zero(a); {int iter; rep(iter, sizeof(a)/sizeof((a)[0])) (a)[iter]=0;} //1次元配列aのゼロ初期化?

typedef long long ll; typedef long double ld;
typedef struct node { int data; struct node *next; } Node;

Node *list[SIZE];
int queue[SIZE]; int q_head = 0, q_tail = 0;

void print_log(); //とりあえず文字を出力してどこまで実行できているか確認(for debug)
void printa_int(int *a, int size); //int型の1次元配列aを出力(for debug)
void printl_int(int size); //リンクリストを出力(for debug)
Node* add_node(int i, int data); void free_list(Node* head);
int queue_empty(); int queue_full();
void enqueue(int data); int dequeue();

int main() {
    int i, N;
    scanf("%d", &N);
    ll *A = array(N, ll);
    rep(i, N) {
        scanf("%lld", A + i);
        if (!A[i]) {
            printf("0\n");
            return 0;
        }
    }
    ll ans = 1, temp = 1;
    rep(i, N) {
        if (ans > 1000000000000000000LL / A[i]) {
            printf("-1\n");
            return 0;
        }
        ans *= A[i];
        temp = ans;
    }
    printf("%lld\n", ans);
    return 0;
}

void print_log() {
    printf("=== print log ===\n");
}
void printa_int(int *a, int size) {
    int i;
    printf("[");
    for (i = 0; i < size; i++)
        printf("%d, ", a[i]);
    printf("]\n");
}
void printl_int(int size) {
    int i;
    Node *t;
    rep(i, size) {
        printf("[%d] ", i);
        for (t = list[i]; t; t = t->next) printf("%d, ", t->data);
        printf("\n");
    }
}
Node* add_node(int i, int data) {
    if (!list[i]) {
        list[i] = (Node*)malloc(sizeof(Node*));
        list[i]->data = data;
    } else {
        Node *t;
        t = list[i];
        list[i] = (Node*)malloc(sizeof(Node*));
        list[i]->data = data;
        list[i]->next = t;
    }
    return list[i];
}
void free_list(Node* head) {
    Node *prev = NULL, *t = head;
    while (t) {
        prev = t;
        t = t->next;
        free(prev);
    }
}
int queue_empty() {
    return q_head == q_tail;
}
int queue_full() {
    return (q_head - 1) % SIZE == q_tail % SIZE;
}
void enqueue(int data) {
    if (queue_full()) {
        printf("full!\n");
        return;
    }
    queue[q_tail++] = data;
    q_tail %= SIZE;
}
int dequeue() {
    if (queue_empty()) {
        printf("empty!\n");
        return -444444;
    }
    int data = queue[q_head++];
    q_head %= SIZE;
    return data;
}