Problem:
Score : 300 points

Problem Statement

Takahashi is organizing a party.

At the party, each guest will receive one or more snack pieces.

Takahashi predicts that the number of guests at this party will be A or B.

Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted.

We assume that a piece cannot be divided and distributed to multiple guests.

Constraints

1 \leq A, B \leq 10^5

A \neq B

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the minimum number of pieces that can be evenly distributed to the guests in both of the cases with A guests and B guests.

Sample Input 1

2 3

Sample Output 1

6

When we have six snack pieces, each guest can take three pieces if we have two guests, and each guest can take two if we have three guests.

Sample Input 2

123 456

Sample Output 2

18696

Sample Input 3

100000 99999

Sample Output 3

9999900000

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;

namespace AtCoder
{
	public class ABC
	{
		static readonly int mod = 1000000007;  // 10^9+7

		static void Main(string[] args)
		{
			Console.SetOut(new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false });


			//int n = int.Parse(Console.ReadLine());
			//string s = Console.ReadLine();
			//string[] s = Console.ReadLine().Split();
			int[] vs = Console.ReadLine().Split().Select(int.Parse).ToArray();
			//long[] a = Console.ReadLine().Split().Select(long.Parse).ToArray();
			int a = vs[0];
			int b = vs[1];
			long ans = LCM(a, b);
			Console.WriteLine(ans);


			Console.Out.Flush();
		}

		#region 最小公倍数 long LCM(long a, long b)
		public static long LCM(long a, long b)
		{
			long gcd;
			if (b == 0) {
				gcd = a;
			} else {
				long x = a, y = b;
				long tmp = x % y;
				while (tmp != 0) {
					x = y;
					y = tmp;
					tmp = x % y;
				}
				gcd = y;
			}
			return a / gcd * b;
		}
		#endregion

	}
}