Problem:
Score : 300 points

Problem Statement

We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.

Initially, the hand of every clock stands still, pointing directly upward.

Now, Dolphin starts all the clocks simultaneously.

In how many seconds will the hand of every clock point directly upward again?

Constraints

1≤N≤100

1≤T_i≤10^{18}

All input values are integers.

The correct answer is at most 10^{18} seconds.

Input

Input is given from Standard Input in the following format:

N
T_1
:
T_N

Output

Print the number of seconds after which the hand of every clock point directly upward again.

Sample Input 1

2
2
3

Sample Output 1

6

We have two clocks. The time when the hand of each clock points upward is as follows:

Clock 1: 2, 4, 6, ... seconds after the beginning

Clock 2: 3, 6, 9, ... seconds after the beginning

Therefore, it takes 6 seconds until the hands of both clocks point directly upward.

Sample Input 2

5
2
5
10
1000000000000000000
1000000000000000000

Sample Output 2

1000000000000000000

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

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

long long int printGCD(long long int v1, long long int v2) {
	long long int a, b, r;

	a = v1;
	b = v2;

	while(b != 0) {
		r = a % b;
		a = b;
		b = r;
	}

	return a;
}
long long int printLCM(long long int v1, long long int v2) {
	long long int a, b, GCD, check1, check2;

	a = v1;
	b = v2;
	GCD = printGCD(a, b);

	while(a > 0) {
		a = a / 10;
		check1++;
	}
	while(b > 0) {
		b = b / 10;
		check2++;
	}
	a = v1;
	b = v2;

	if(check1 + check2 >= 18)
		return 1000000000000000000LL;
	
	return (long long int)(a * b / GCD);
}

int main(void) {
	long long int array[100];
	int N, i;

	scanf("%d", &N);
	for(i = 0; i < N; i++) {
		scanf("%lld", &array[i]);
	}

	for(i = 0; i < N-1; i++) {
		array[i+1] = printLCM(array[i], array[i+1]);
	}

	printf("%lld", array[i]);

	return EXIT_SUCCESS;
}