Problem:
Score : 100 points

Problem Statement

You are given positive integers X and Y.
If there exists a positive integer not greater than 10^{18} that is a multiple of X but not a multiple of Y, choose one such integer and print it.
If it does not exist, print -1.

Constraints

1 ≤ X,Y ≤ 10^9

X and Y are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

Print a positive integer not greater than 10^{18} that is a multiple of X but not a multiple of Y, or print -1 if it does not exist.

Sample Input 1

8 6

Sample Output 1

16

For example, 16 is a multiple of 8 but not a multiple of 6.

Sample Input 2

3 3

Sample Output 2

-1

A multiple of 3 is a multiple of 3.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<math.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>
#include<set>
#include<vector>
#include<string.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <db, db> pdd;

const int maxn = 2005;
const int Mod=1000000007;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const db PI = acos(-1);
const db ERR = 1e-10;
const ll mm=1e18;
#define Se second
#define Fi first

int main()
{
    ios::sync_with_stdio(false);
    ll x,y;
    cin>>x>>y;
    if(x%y==0)
    {
        cout<<-1<<endl;
        return 0;
    }
    for(ll i=x;i<=mm;i+=x)
    {
        if(i%y!=0)
        {
            cout<<i<<endl;
            return 0;
        }
    }
    return 0;
}