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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
#include<stdio.h>
#include <cstdint>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define all(a) a.begin(), a.end()
#define fi first
#define se second
#define RNG(x, a, n) &((x)[a]), &((x)[n])
typedef pair<int, int> PII;
typedef long long ll;
template<typename T> bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; } return false; }
template<typename T> bool chmin(T& a, const T& b) { if (a > b) { a = b; return true; } return false; }
const int inf = 1e9 + 7;
const ll longinf = 1LL << 60;
const ll mod = 1e9 + 7;


ll a, b, ans;

long long GCD(ll a, ll b) {
	if (b == 0) return a;
	else return GCD(b, a % b);
}


int main() {

	cin >> a >> b;

	ans = a * b / GCD(a, b);
	
	cout << ans << endl;

	return 0;
}