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 <algorithm>
#include <bitset>
#include <cassert>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>

using namespace std;
using Graph = vector<vector<int>>;
typedef long long ll;
typedef pair<ll, ll> P;
inline ll min(ll x, ll y) { return x < y ? x : y; }
inline ll max(ll x, ll y) { return x > y ? x : y; }

#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i < (int)(n); i++)
#define all(c) c.begin(), c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << (x) << endl
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)

ll gcd(ll a, ll b) {
    if (b == 0) {
        return a;
    }

    return gcd(b, a % b);
}

int main() {
    ll A, B;
    cin >> A >> B;
    cout << A * B / gcd(A, B);
    return 0;
}