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.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

  public static void main(String[] args) {
    InputReader in = new InputReader(System.in);
    PrintWriter out = new PrintWriter(System.out);

    out.println(lcm(in.nextLong(), in.nextLong()));

    out.close();
  }

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

  static long gcd(long a, long b) {
    if (a == 0)
      return b;
    return gcd(b % a, a);
  }

  static class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
      reader = new BufferedReader(new InputStreamReader(stream), 32768);
      tokenizer = null;
    }

    public String next() {
      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          tokenizer = new StringTokenizer(reader.readLine());
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }
      return tokenizer.nextToken();
    }

    public int nextInt() {
      return Integer.parseInt(next());
    }

    public long nextLong() {
      return Long.parseLong(next());
    }

    public double nextDouble() {
      return Double.parseDouble(next());
    }
  }

  public static int[][] lexSort(int[][] a) {
    if (a.length <= 1)
      return a;

    int mid = a.length / 2;
    int[][] left = new int[mid][2];
    int[][] right = new int[a.length - mid][2];

    for (int i = 0; i < mid; ++i)
      left[i] = a[i];

    for (int i = mid; i < a.length; ++i)
      right[i - mid] = a[i];

    return lexMerge(lexSort(left), lexSort(right));
  }

  public static int[][] lexMerge (int[][] l, int[][] r) {
    int[][] merged = new int[l.length + r.length][2];

    int i = 0, j = 0, k = 0;
    while (i < l.length && j < r.length) {
      if (l[i][0] < r[j][0] || l[i][0] == r[j][0] && l[i][1] <= r[j][1])
        merged[k++] = l[i++];
      else
        merged[k++] = r[j++];
    }

    while (i < l.length)
      merged[k++] = l[i++];
    while (j < r.length)
      merged[k++] = r[j++];

    return merged;
  }
}