D: Is greed the best?

story

In Japan, where there are 1, 5, 10, 50, 100, 500 yen coins, it is known that the number of coins can be minimized by using as many coins as possible when paying a certain amount. ..

If the amount of coins is different from that of Japan, it is not always possible to minimize it by paying greedily.

What are the conditions that the amount of coins must meet in order to be optimally paid greedily?

problem

TAB was curious about the above, so I decided to first consider the case where there are only three types of coins, 1, A and B.

Since A and B are given, output the smallest amount of money that will not minimize the number of sheets if you pay greedily.

Also, if the greedy algorithm is optimal for any amount of money, output -1.

Input format


A B

Constraint

* 1 <A \ leq 10 ^ 5
* A <B \ leq 10 ^ 9



Input example 1


4 6

Output example 1


8

If you greedily pay 8 yen, you will pay 6 + 1 \ times 2 for a total of 3 cards, but you can pay 4 \ times 2 for a total of 2 cards.

Input example 2


2 1000000000

Output example 2


-1

It is best to greedily pay any amount.





Example

Input

4 6


Output

8
//#pragma once
#include <stdio.h>
#include <stdlib.h>
#include <string>
#include <iostream>
#include <queue>
#include <algorithm>
#include <sstream>
#include <vector>
#include <math.h>
#include <set>
#include <map>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cctype>
#include <cstdlib>      // srand,rand


using namespace std;
#define modd 1000000007



typedef pair<long long, long long> pl;
typedef pair<double, long long> pdl;



typedef string::const_iterator State;
class ParseError {};




#define ll long long




class UnionFind {
public:
	vector <ll> par;
	vector <ll> siz;

	// Constructor
	UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL){
		for (ll i = 0; i < sz_; ++i) par[i] = i;
	}
	void init(ll sz_) {
		par.resize(sz_);
		siz.assign(sz_, 1LL);  // resize だとなぜか初期化されなかった
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}

	// Member Function
	// Find
	ll root(ll x) { // 根の検索
		while (par[x] != x) {
			x = par[x] = par[par[x]]; // x の親の親を x の親とする
		}
		return x;
	}

	// Union(Unite, Merge)
	bool merge(ll x, ll y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;

		// merge technique（データ構造をマージするテク．小を大にくっつける）
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		return true;
	}



	bool issame(ll x, ll y) { // 連結判定
		return root(x) == root(y);
	}

	ll size(ll x) { // 素集合のサイズ
		return siz[root(x)];
	}


};


class SegmentTree {
private:
	int n;
	vector <ll> node;

public:
	// 元の配列　v をセグメントツリーで表現する。
	SegmentTree(vector<ll> v) {
		//最下段のノード数は元配列のサイズ以上になる最小の2冪->これをnとおく。
		//セグ木全体で必要なノード数は2n-1個である。
		int sz = v.size();
		n = 1; 
		while (n < sz)n *= 2;

	}

};




long long N, M, K, a, b, c, d, e, H, W, L,T;
ll A[300005] = {};
ll B[300005] = {};
ll C[100][3005] = {};
bool f, ff;
string S,SS;
set <long long>sll;
pl bufpl;
vector <long long>vl[100005];
vector <long long>vll;
vector <pl>vpl;
vector <string> vs;
set<ll> llset[50];
multiset<ll> llmset;
queue<ll> ql;


struct ST
{
	ll first;
	ll second;
	ll kaisuu;
	

	bool operator<(const ST& another) const
	{
		return first < another.first;//比較
	};

};



/*vector <ST> vst;
ST st[200005];
ST bufst;
bitset<5000> bits;*/


long long modinv(long long aa, long long mm) {
	long long bb = mm, uu = 1, vv = 0;
	while (bb) {
		long long tt = aa / bb;
		aa -= tt * bb; swap(aa, bb);
		uu -= tt * vv; swap(uu, vv);
	}
	uu %= mm;
	if (uu < 0) uu += mm;
	return uu;
}


ll minn, maxx;
double ansa;

ll zettai(ll aa) {

	if (aa < 0) {
		aa *= -1;
	}
	return aa;

}



double ab[200];

pdl bufdl;
vector<pdl> vdl;

int main() {
	
	cin >> a>>b;
	
	if (b % a == 0) {
		cout << -1 << endl;
	}
	else {
		f = true;
		M = b / a;
		M++;

		for (ll i = M; i <= M+100000; i++) {
			N = i * a;

			e = N / b;
			H = (N - e * b) / a;
			d = e + H+(N-e*b-H*a);

			if (i<d) {
				f = false;
				break;
			}
		}

		if (f) {

			cout << -1 << endl;

		}
		else {
			cout << N << endl;
		}
	}

	

	//cout << fixed << setprecision(10) << ansa << endl;
	return 0;



}

