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>
using namespace std;

typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll,ll> pll;
typedef vector<bool> vb;
const ll oo = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-9;
#define sz(c) ll((c).size())
#define all(c) begin(c), end(c)
#define FOR(i,a,b) for (ll i = (a); i < (b); i++)
#define FORD(i,a,b) for (ll i = (b)-1; i >= (a); i--)
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define eb emplace_back
#define xx first
#define yy second
#define has(c,i) ((c).find(i) != end(c))
#define DBGDO(X) ({ if(1) cerr << "DBGDO: " << (#X) << " = " << (X) << endl; })

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	ll x, y;
	cin >> x >> y;
	cout << (x%y ? x : -1) << endl;

}