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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayDeque;
import java.util.Queue;
import java.util.Random;
import java.util.Scanner;
 


public class Main {
	private static Scanner sc = new Scanner(System.in);
	//result = Math.pow(num1, num3)
	//StringBuffer str = new StringBuffer(hoge1);
	//String hoge2 = str.reverse().toString();
	//map.containsKey(A)

	//Map<String, Integer> map = new HashMap<String, Integer>(n);
	/*for ( キーのデータ型 key : マップの名前.keySet() ) {
		データのデータ型 data = マップの名前.get( key );
		
		// keyやdataを使った処理;
	}*/
	//int i = Integer.parseInt(s);
	//Queue<String> qq=new ArrayDeque<>(); //add,poll,peek
	//Deque<String> qq=new ArrayDeque<>();//push,pop,peek
	//ArrayList<String> cc = new ArrayList<>(n);
	//Collections.sort(list);
	//Array.sort(list);   cc.contains(tmp)
	//Arrays.asList(c).contains("a")
	//list.set(1,"walk");//1 1 2 3 5
	//static long mod =1000000007;
	//static 	ArrayList<Integer> cc = new ArrayList<>(100);
	
	
	
	
	public static void main(String[] args) {
		//int N=sc.nextInt();for(int i=0;i<N;i++) {}
		//int a[]=new int[N];
		//Ae2+Be2=Ce2
		int N=sc.nextInt();
		long T[]=new long[N];
		long ll=1;
		for(int i=0;i<N;i++) {
			
			T[i]=sc.nextLong();
			ll=lcm(ll,T[i]);
			
		}
		p(ll);
		}
		
		
	
	
	
	public static long lcm(long a, long b) {
        return a / gcd(a, b) * b;
	}
	
	public static long gcd(long a, long b) {
        return b == 0 ? a: gcd(b, a % b);
	}

	


	
	static void p(String ans) {System.out.println(ans);};
	static void p(int ans) {System.out.println(ans);};
	static void p() {System.out.println();};
	static void p(long ans) {System.out.println(ans);};
	static void p(double ans) {System.out.println(ans);};
	
}