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: Wrong Answer)

/*
* @Author Silviase(@silviasetitech)
* For ProCon
*/

import java.util.*;
import java.lang.*;
import java.math.*;

class Main{

    static int n;
    static long[] times;
    static int m;
    static long ans;
    static int[] rate;
    static int var;
    static int ansmin;
    static int ansmax;
    static int[] w;
    static int[] ww;
    static boolean[] visit;
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        
        n = sc.nextInt();
        times = new long[n];
        for (int i = 0; i < n; i++) {
            times[i] = sc.nextLong();
        }
        for (int i = 0; i < n; i++) {
            if (i == 0) {
                ans = times[i];
            }else{
                ans = times[i]/gcd(times[i], times[i-1])*times[i-1];
            }
        }
        System.out.println(ans);

        sc.close();
    }

    public static long gcd(long a, long b) {
        
        if (a < b) {
            long tmpp = a;
            a = b;
            b = tmpp;
        }
        if (b == 0) {
            return a;
        }else{
            return gcd(b, a%b);
        }
    }

    public static void dfs(int placenow) {
        // if went all -> success! 
        // if not went all -> fail...
        
        /*
         dfs
         Go All Place that there is way to and not having been yet. 

         if island 1 is start point, dfs(1);
         if there is way to island 2 and island 3,
         - island 2 changes non-visited -> visited, and dfs(2).
         - island 3 changes non-visited -> visited, and dfs(3).
         

        */

        visit[placenow] = true;
        boolean success = true;
        for (int i = 0; i < n; i++) {
            if (visit[i] == false) { // not go all place
                success = false;
                break;
            }
        }
        if (success) {
            ans++;
            visit[placenow] = false;
            return;
        }

        for (int i = 0; i < m; i++) {
            if (w[i] == placenow && visit[ww[i]] == false ) {
                dfs(ww[i]);
            }else if(ww[i] == placenow && visit[w[i]] == false){
                dfs(w[i]);
            }else{
                continue;
            }
        }
        visit[placenow] = false;
        return;
    }

}