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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, c) for (int i = 0; i < (int)c; i++)
using ll = long long;

// 最大公約数(GCD:greatest common divisor) ユークリッドの互除法
ll gcd(ll a, ll b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}

// 最小公倍数(LCM:least common multiple)
// 公式 a * b = gcd(a, b) * lcm(a, b)
// a * b / gcd(a, b) = lcm(a, b)
// a, bはそれぞれ10^18なのでa*bするとTLEしちゃう。なので、 a / gcd(a,b)してから*b
ll lcm(ll a, ll b) {
    return (a / gcd(a, b)) * b;
}

int main() {
    int n;
    cin >> n;
    ll t, ans = 1;
    rep(i, n) {
        cin >> t;
        ans = lcm(ans, t); //入力の度に最小公倍数を探す
    }

    cout << ans << endl;
    return 0;
}