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 Ruby (2.3.3) (judge verdict: Accepted)

def gcd_of_two(a, b)
  small, large = [a, b].sort
  if small == 0
    return large
  else
    gcd_of_two(small, large % small)
  end
end

def gcd_of_many(array)
  current = array.shift
  until array.empty?
    following = array.shift
    current = gcd_of_two(current, following)
  end
  return current
end

def lcm_of_two(a, b)
  gcd = gcd_of_two(a, b)
  return a * b / gcd
end

def lcm_of_many(array)
  current = array.shift
  until array.empty?
    following = array.shift
    current = lcm_of_two(current, following)
  end
  return current
end

_n = gets.to_i
ts = readlines.map(&:to_i)
puts lcm_of_many(ts)