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 JavaScript (node.js v5.12) (judge verdict: Runtime Error)

'use strict';

const input = require('fs').readFileSync('/dev/stdin', 'utf8').trim().split('\n');
//　最初の行の数値を獲得
const info = input.shift();

// 配列の重複を消したかったので
const inputSet = new Set(input);

// 全部掛け合わせたものを算出
let mul = 1;

for(let i of inputSet) {
	mul *= i;
}

// 最大公約数の式
let gcd = (a,b) =>{
	return a == 0 ? b : gcd(b % a , a);
}

// 最小公倍数を求める式
let lcm = (c,d) =>{
	return c * d / gcd(c, d);
}

// setを配列に戻す
let nums = [...inputSet];
let lcmNum = 1;

// 片っ端から、最小公倍数を取っていく
for(let i of nums){
  lcmNum = lcm(lcmNum,i);
}

console.log(lcmNum);