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>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int min(int a, int b) {
    if (a <= b) {
        return a;
    } else {
        return b;
    }
}

int main(void) {
    long long int a, b, gcd = 1, sqrt;
    scanf("%lld%lld", &a,&b);
    long long int ans;
    sqrt = (long long int)pow(min(a,b), 0.5)+1;
    for (long long int i = 2; i <= sqrt; i++) {
        if (a%i==0 && b%i==0) {
            gcd = i;
        }
    }
    ans = a*b/gcd;
    printf("%lld\n", ans);
    return 0;
}