A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
题意：你有一辆车，油箱可装L升油，现在，你要从0出发走到点a（在一位数轴上），每走一个单位，需要消耗1升油。在0
到a之间，有且只有一个加油站，位于f。从0开到a或者从a开到0算做一个journey。你需要走k个journey。问，你最少需要加几次油，如果不可以完成k次journey，输出-1。
思路：每到一次加油站，我们都需要进行判断，如果还要走一个journey.的话，油箱里的油是否足够再次走到加油站，如果不用再走一个journey.的话，只需判断油箱里的油是否可以从加油站走到0或着a点。如果L小于0的话，说明走不到目的地，那么直接退出。
#include "iostream"
using namespace std;
int main()
{
int a,b,f,k,ans=0,L,flag=0;
cin>>a>>b>>f>>k;
L=b;
while(k--){
if(L<f){flag=1;break;}
else L-=f;
if(k){if(L<(a-f)*2) ans++,L=b;}
else{
if(L<(a-f)){
ans++,L=b;
if(L<(a-f)) flag=1;
}
break;
}
L-=(a-f)*2;
if(L<0) {flag=1;break;}
k--;
if(k){if(L<(f-0)*2) ans++,L=b;}
else if(L<(f-0)) ans++,L=b;
L-=f;
if(L<0) {flag=1;break;}
}
if(flag) cout<<"-1"<<endl;
else cout<<ans<<endl;
}