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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Text;
using System.Linq;
using System.Collections;
using System.Collections.Generic;
using static System.Console;
using static System.Math;

namespace NotFounds
{
    public class Program
    {
        public static void Main(string[] args)
        {
            new Program().Solve();
        }

        public void Solve()
        {
            var cin = new MyInputStream();
            var n = cin.ReadInt() - 1;
            var x = cin.ReadLong();
            for (int i = 0; i < n; ++i)
            {
                var t = cin.ReadLong();
                x = Lcm(x, t);
            }
            WriteLine(x);
        }

        private static void PrintYN(bool b)
        {
            //WriteLine(b ? "YES" : "NO");
            WriteLine(b ? "Yes" : "No");
        }

        private static string Reverse(string s)
        {
            return string.Join("", s.Reverse());
        }

        /// <summary>最大公約数</summary>
        public static long Gcd(long m, long n)
        {
            if (n == 0) return m;
            return Gcd(n, m % n);
        }

        /// <summary>最小公倍数</summary>
        public static long Lcm(long m, long n)
        {
            return m / Gcd(m, n) * n;
        }
    }

    public class MyInputStream
    {
        private char separator = ' ';
        private Queue<string> inputStream;
        public MyInputStream(char separator = ' ')
        {
            this.separator = separator;
            inputStream = new Queue<string>();
        }

        public string Read()
        {
            if (inputStream.Count != 0) return inputStream.Dequeue();
            string[] tmp = Console.ReadLine().Split(separator);
            for (int i = 0; i < tmp.Length; ++i)
                inputStream.Enqueue(tmp[i]);
            return inputStream.Dequeue();
        }
        public string ReadLine() { return Console.ReadLine(); }
        public int ReadInt() { return int.Parse(Read()); }
        public long ReadLong() { return long.Parse(Read()); }
        public double ReadDouble() { return double.Parse(Read()); }
        public string[] ReadStrArray(long N) { var ret = new string[N]; for (long i = 0; i < N; ++i) ret[i] = Read(); return ret;}
        public int[] ReadIntArray(long N) { var ret = new int[N]; for (long i = 0; i < N; ++i) ret[i] = ReadInt(); return ret;}
        public long[] ReadLongArray(long N) { var ret = new long[N]; for (long i = 0; i < N; ++i) ret[i] = ReadLong(); return ret;}
    }
}