CodeForces

C. Bus

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

      昨天做的测试里的题，水的没话说，但是栽在上面了。题意是有一条坐标0-a的路，f（0-a之间）为加油点，车从0出发，最多装b升油，油耗为一米一单位。让车来回走，0-a，a-0都算一趟，求k趟最少加油次数。数据不多，直接模拟就是了，没有任何算法。自以为是的加上了句话，结果一直不对一直不对，在代码里留下了。

代码如下：

#include <iostream>using namespace std;int main(){    int a, b, f, k;    cin>>a>>b>>f>>k;    /*              //本来自以为是的加上的剪枝，结果就错在了这里。。    if(b<f*2 || b<(a-f)*2)    {        cout<<"-1"<<endl;        return 0;    }*/    int len[10010];    len[0]=f;    for(int i=1; i<k; i++)    {        if(i%2==0)            len[i]=f*2;        else            len[i]=(a-f)*2;    }    if(k%2==1)        len[k]=a-f;    else        len[k]=f;    int oil=b;    int ans=0;    for(int i=0; i<=k; i++)    {        if(oil<len[i])        {            ans=-1;            break;        }        if(i<k && len[i]+len[i+1]>oil)        {            oil=b;            ans++;        }        else if(i<k && len[i]+len[i+1]<=oil)        {            oil-=len[i];        }    }    cout<<ans<<endl;    return 0;}