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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        C solver = new C();
        solver.solve(1, in, out);
        out.close();
    }

    static class C {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            // 整数の入力
            long a = Long.parseLong(in.next());
            long b = Long.parseLong(in.next());
            long ans = 0;

            ans = lcm(a, b);

            // 出力
            out.println(ans);

        }

        static long lcm(long a, long b) {
            long temp;
            long c = a;
            c *= b;
            while ((temp = a % b) != 0) {
                a = b;
                b = temp;
            }
            return (long) (c / b);
        }

    }
}