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: Wrong Answer)

#include<stdio.h>
#define BUFSIZE 10000
int main(void){
	int reminder[BUFSIZE]={0};
	long a,b;
	int i;
	int m=0;
    scanf("%d %d",&a,&b);
    for(i=1;i<=a;i++){
		if(a%i==0){
			reminder[m]=i;
							
			m++;
		}
	}
	int n=0;
	int N_reminder[BUFSIZE]={0};
		for(int m=0;m<BUFSIZE;m++){
		    if(reminder[m]==0){
				break;
			}    
				if(b%reminder[m]==0){
					N_reminder[n]=reminder[m];
					n++;
				}
		}
long gcd;		
for(int j=0;j<BUFSIZE;j++){
	if(N_reminder[j]==0){
		gcd=N_reminder[j-1];
		break;
	}
}
printf("%d\n",gcd);
printf("%d",a/gcd*b/gcd*gcd);			
		
			

}