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)

import numpy as np

# 2整数の最大公約数を求める関数を定義する
def get_gcd(a, b):
    a, b = max(a, b), min(a, b)
    c = a % b
    if c == 0:
        return b
    else:
        return get_gcd(b, c)

# --------------------------------------------

# 時計の数を入力する
N = int(input())
# 時計の周回時間を入力する
arr_T = np.empty(N, dtype = int)
for i in range(N):
    arr_T[i] = int(input())
# 周回時間を降順に並べる
arr_sort_T = np.sort(arr_T)[::-1]

# 全ての周回時間の最小公倍数を求める
# 最大周回時間を取り出す
max_T = np.max(arr_T)
m = max_T
if N != 1:
    # 繰り返し
    for n in arr_sort_T[1:]:
        gcd = get_gcd(m, n)
        m = (m // gcd) * n

# 結果を出力する
print(m)