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
struct Fast {Fast(){std::cin.tie(0);ios::sync_with_stdio(false);}} fast;

/* short */
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define ALL(v) begin(v), end(v)

/* REPmacro */
#define FOR(i, s, n) for (int i = (s); i < (n); i++)
#define RFOR(i, s, n) for (int i = (s); i >= (n); i--)
#define rep(i, n) FOR(i, 0, n)
#define repi(i, n) FOR(i, 1, n + 1)
#define rrep(i, n) RFOR(i, n - 1, 0)
#define rrepi(i, n) RFOR(i, n, 1)

/* debug */
#define pp(v) cerr << #v "=" << (v) << endl;

/* alias */
using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using pii = pair<int, int>;
using vs = vector<string>;
using Graph = vvi;
template <typename T> using PQ = priority_queue<T>;
template <typename T> using minPQ = priority_queue<T, vector<T>, greater<T>>;

/* iostream */
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }

/* const */
const int INF = 1001001001;
const ll LINF = 1001001001001001001ll;
const int MOD = 1e9 + 7;
const int dx[] = {0, 1, 0, -1, 1, -1, 1, -1}, dy[] = {1, 0, -1, 0, 1, -1, -1, 1};

/* func */
#define UNIQUE(v) v.erase( unique(ALL(v)), v.end() );
#define TIME system("date +%M:%S.%N")
inline bool inside(int y, int x, int H, int W) {return y >= 0 && x >= 0 && y < H && x < W;}
template <typename T> inline bool chmin(T& a, const T& b) {if (a > b) a = b; return a > b;}
template <typename T> inline bool chmax(T& a, const T& b) {if (a < b) a = b; return a < b;}
template<class T> T gcd(const T &a, const T &b) { return a < b ? gcd(b, a) : b ? gcd(b, a % b) : a; }
template<class T> T lcm(const T &a, const T &b) { return a / gcd(a, b) * b; }
template<class T> T div_ceil(const T &a, const T &b) { return (a + b - 1) / b; }

signed main() {
  int n;cin>>n;
  vi t(n);cin>>t;
  int ans = 1;
  rep(i,n) ans = lcm(ans,t[i]);
  cout << ans << endl;
}