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 <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <queue>

using namespace std;

typedef long long ll;
typedef vector<int> vint;
typedef vector<vector<int> > vvint;
typedef vector<long long> vll, vLL;
typedef vector<vector<long long> > vvll, vvLL;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<n;++i)
#define mod (ll)(1e9+7)
#define FIX(a) ((a)%mod+mod)%mod
#define ALL(obj) (obj).begin(), (obj).end()
#define rALL(obj) (obj).rbegin(), (obj).rend()
#define INF 1000000000 //1e9
#define LLINF 2000000000000000000LL //2e18
#define fi first
#define se second
#define pb push_back
int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll x,y;
    cin >> x >> y;
    if(x<y){
        cout << x << endl;
    }else if(x%y==0){
        cout << -1 << endl;
    }else{
        cout << x << endl;
    }
    

    return 0;
}