Codeforces Round #436 (Div. 2)

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.

题意：

x轴上0-a长，f点上有一个加油站，b为车的油库，现在车要走k次，从0到n或者从n到0为一次。求最少加油次数。

POINT：

贪心，每次看一看这次要不要加油，如果不加油跑不到下一个加油站，就加，不然就不加。当然最后第k次稍微不同一点，

我把来回都拉成一条线了。代码可能比较抽象。

#include <iostream>#include <string.h>#include <stdio.h>#include <math.h>#include<algorithm>using namespace std;#define LL long longint cnt[222];int main(){    int a,b,f,k;    scanf("%d %d %d %d",&a,&b,&f,&k);    int now=1;    int you=b;    int flag=1;    int cnt=0;    if(b<f||b<(a-f)) flag=0;    if(k>=2&&b<2*(a-f)) flag=0;    if(k>=3&&b<2*f) flag=0;    if(!flag){        printf("-1\n");        return 0;    }    int aim=a-f;    while(now!=k+1){        you-=a-aim;        if(now==k){            if(you<aim){                cnt++;            }            break;        }        else{            if(you>=2*aim){                you=you-aim;                now++;            }            else{                you=b-aim;                cnt++;                now++;            }            aim=a-aim;        }    }    printf("%d\n",cnt);}