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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = mod * mod;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

//繰り返し二乗法
ll mod_pow(ll a, ll n, ll m) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%m;
		a = a * a%m; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

//逆元(Eucledean algorithm)
ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 18;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}
using mP = pair<modint, modint>;

int dx[4] = { 0,1,0,-1 };
int dy[4] = { 1,0,-1,0 };

struct SuffixArray {
	struct SAComp {
		const int h;
		const vector<int> &g;
		SAComp(int h, vector<int> &g) : h(h), g(g) { ; }
		bool operator() (int a, int b) {
			return a != b && (g[a] != g[b] ? g[a] < g[b] : g[a + h] < g[b + h]);
		}
	};
	int n;
	string str;
	vector<int> sa, lcp;
	SuffixArray(const string &t) : n(t.size()), sa(n + 1), lcp(n + 1) {
		str = t;
		// build SA
		vector<int> g(n + 1, 0), b(n + 1, 0);
		rep(i, n + 1) { sa[i] = i; g[i] = str[i]; }
		sort(begin(sa), end(sa), SAComp(0, g));
		for (int h = 1; b[n] != n; h *= 2) {
			SAComp comp(h, g);
			sort(sa.begin(), sa.end(), comp);
			rep(i, n) b[i + 1] = b[i] + comp(sa[i], sa[i + 1]);
			rep(i, n + 1) g[sa[i]] = b[i];
		}
		// build LCP
		int h = 0;
		rep(i, n + 1) b[sa[i]] = i;
		rep(i, n + 1) {
			if (b[i]) {
				for (int j = sa[b[i] - 1]; j + h<n && i + h<n && str[j + h] == str[i + h]; ++h);
				lcp[b[i]] = h;
			}
			else {
				lcp[b[i]] = -1;
			}
			if (h > 0) --h;
		}
	}
	~SuffixArray() { str.clear(); }
	/*int find(string t) {
	int m = t.size();
	string p;
	p = t;
	int left = -1, right = n + 1;
	while (left + 1 < right) {
	int mid = (left + right) / 2;
	if (strncmp(str + sa[mid], p, m) < 0) left = mid;
	else right = mid;
	}
	return strncmp(str + sa[right], p, m) == 0 ? sa[right] : -1;
	}*/
};


ll gcd(ll a, ll b)
{
   if (a%b == 0)
   {
       return(b);
   }
   else
   {
       return(gcd(b, a%b));
   }
}

ll lcm(ll a, ll b)
{
   return a / gcd(a, b) * b;
}

void solve() {
  int N; cin >> N;
  vector<ll> T(N);
  rep(i, N){
    cin >> T[i];
  }
  
  if(N == 1){
    cout << T[0] << endl;
    return;
  }
  ll ini = lcm(T[0], T[1]);
  ll ans=ini;
  for(int i=2; i<N; i++){
    ans = lcm(ans, T[i]);
  }
  
  cout << ans << endl;
}

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  //cout << fixed << setprecision(10);
  //init_f();
  //init();
  //int t; cin >> t; rep(i, t)solve();
  solve();
//   stop
    return 0;
}