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)

//
//AtCoder Beginner Contest 070
//C
//

#include <iostream>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <queue>
#include <map>
#include <set>
#include <tuple>
#include <functional>
#include <cassert>
#include <bitset>

using namespace std;

#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define REPD(i,n) for(ll i=n-1;i>=0;i--)
#define FOR(i,a,b) for(ll i=a;i<=(ll)(b);i++)
#define FORD(i,a,b) for(ll i=a;i>=(ll)(b);i--)
#define ALL(x) (x).begin(),(x).end()
#define RALL(x) (x).rbegin(),(x).rend()
#define SIZE(x) (int)(x).size()
#define SIZEL(x) (ll)(x).size()
#define MSG(a) cout << #a << " " << a << endl;

using ll = long long;
using ld = long double;
using Graph = vector<vector<int>>;

const double PI = 3.14159265358979323846;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

ll gcd(ll x, ll y) { return (x % y)? gcd(y, x % y): y; }    //最大公約数
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }            //最小公倍数

int main()
{
    int n;
    cin >> n;

    vector<ll> t(n);
    REP(i, n) cin >> t[i];

    ll ans = 1;

    REP(i, n){
        ans = lcm(ans, t[i]);
    }
    cout << ans << endl;
    return 0;
}