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>
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define erep(i,a,b) for(int i=a;i<=(int)(b);++i)
#define per(i,a,b) for(int i=(b);i>(a);--i)
#define eper(i,a,b) for(int i=((int)(a));i>=b;--i)
#define pb push_back
#define mp make_pair
#define INF (1<<28)-1
#define MOD 1000000007
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef pair<int,int> Pii;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0};
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a, b)*b;}

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