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 Python (3.4.3) (judge verdict: Accepted)

from statistics import mean, median,variance,stdev
import numpy as np
import sys
import math
import fractions
import itertools
import copy
import collections
from operator import itemgetter
#以下てんぷら
def j(q):
    if q==1: print("Yes")
    else:print("No")
    exit(0)


def ct(x,y):
    if (x>y):print("+")
    elif (x<y): print("-")
    else: print("?")

def ip():
    return int(input())
def printrow(a):
    for i in range(len(a)):
        print(a[i])
def combinations(n,r):
    if n<r:return 0
    return math.factorial(n) // (math.factorial(n - r) * math.factorial(r))
def permutations(n,r):
    if n<r:return 0
    return math.factorial(n) // math.factorial(n - r)
def lcm(x, y):
    return (x * y) // fractions.gcd(x, y)
#n = ip()                                     #入力整数1つ
#n,k= (int(i) for i in input().split())       #入力整数横2つ以上
#a = [int(i) for i in input().split()]        #入力整数配列
#a = input()                                  #入力文字列
#a = input().split()                          #入力文字配列
n = ip()                      #入力セット(整数改行あり)(1/2)
a=[ip() for i in range(n)]    #入力セット(整数改行あり)(2/2)
#a=[input() for i in range(n)]    #入力セット(整数改行あり)(2/2)
#jの変数はしようできないので注意
#全足しにsum変数使用はsum関数使用できないので注意
#こっから本文
s = a[0]
for i in range(1,n):
    s = lcm(s,a[i])
print(s)