// #pragma GCC optimize(2)
/* // 快读：返回读入的第一个整数
inline int read()
{
 int x=0,f=1;char ch=getchar();
 while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
 while (ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}
 return x*f;
}*/
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <stack>
#include <map>
#include <queue>
#include <cmath>
#include <cstdio>
#include <deque>
#include <unordered_map>
#include <unordered_set>
#include <cfloat>
#include <set>
#include <climits>
using namespace std;
using f64 = long double;
// #define int long long
// # define int unsigned long long
// #define endl '\n'
// #define minn(a,b) (a>b?b:a)
const int N = 1e5 + 10, M = N * N;
const int base = 1e9;
const int basebit = 9;
const int mod = 1e9 + 7;
const long long INF = 1e18;
typedef long long ll;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<pair<int, int>, int> PIII;
typedef pair<f64, int> PDI;

int x, y, a, b;
void solve()
{
	// 有个问题，，船一定要装满吗？
	cin >> x >> y >> a >> b;
	// 直接模拟
	int xx = 0, yy = 0;
	int tim = 0, size = 0; // size为0时代表在本岸，为1时在对岸
	while (1)			   // 如果此时在本岸且羊的数量可以一条船带走，结束模拟
	{
		if (!size)
		{
			int tem1 = max(0, y - (x + b + 1)); // 必须带走的狼群数量
			if (tem1 > a)
			{
				cout << "-1" << endl;
				return;
			}
			int tem = (a - tem1) / 2;
			y -= tem1 + tem;
			yy += tem1 + tem;
			x -= tem;
			xx += tem;
		}
		else
		{
			int tem1 = max(0, y - (x + b + 1)); // 必须带走的狼群数量
			if (tem1 > a)
			{
				cout << "-1" << endl;
				return;
			}
			yy -= tem1;
			y += tem1;
		}
		tim++, size ^= 1;
		if (x <= a && size == 0)
			break;
		if (tim >= 1000)
		{
			cout << "-1" << endl;
			return;
		}
	}
	tim++;
	cout << tim << endl;
}
signed main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int t = 1;
	// cin >> t;
	while (t--)
		solve();
	return 0;
}