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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#define max(a, b)	(((a) > (b)) ? (a) : (b))		/* ２個の値の最大値 */
#define min(a, b)	(((a) < (b)) ? (a) : (b))		/* ２個の値の最小値 */
#define ENTER		printf("\n")					/* 改行プリント */
int DBG = 1;										/* デバッグプリント 提出時は0 */

long gcd(long m, long n);							/* 最大公約数 */
long lcm(long m, long n);							/* 最小公倍数 */

/* 最大公約数 */
long gcd(long m, long n) {
	long temp;
	while (m % n != 0)
	{
		temp = n;
		n = m % n;
		m = temp;
	}
	return n;
}

/* 最小公倍数 */
long lcm(long m, long n) {
	return m*n/gcd(m,n);
}
/* main *************************************************************/
int main()
{
	int		a,b;
	long    ans;

	
	scanf("%d",&a);
	scanf("%d",&b);
	
	ans = lcm( (long)a, (long)b);
	
	printf("%lld\n",ans);


	return 0;
}