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.

Examples

Input

2
2
3


Output

6


Input

5
2
5
10
1000000000000000000
1000000000000000000


Output

1000000000000000000


import java.util.*;

public class Main {
	//static boolean f;
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		/* 入力の受け取り
		 * String s = sc.next();
		 * int a = sc.nextInt();
		 * long a = sc.nextLong();
		 */
		
		// 文字列(String)を文字(char)に分解
		// char[] s_char = s.toCharArray();
		
		int n = sc.nextInt();
		long t[] = new long[n];
		for(int i=0;i<n;i++)
			t[i] = sc.nextLong();
		
		long time = t[0];
		
		if(n != 1) {
			for(int i=1;i<n;i++)
				time = lcm(time,t[i]);
		}
		
		System.out.println(time);
		
		/*
		 * if(f)
		 *	System.out.println("Yes");
		 * else
		 * 	System.out.println("No");
		 */
	}
	
	// ユークリッドの互除法(最大公約数を求める)
	static long gcd(long x, long y) {
		return x%y==0? y : gcd(y,x%y);
	}
		
	// 最小公倍数を求める
	static long lcm(long x, long y) {
		return x / gcd(x, y) * y;
	}
		
}
