CF1185A.Ropewalkers

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题目描述

Polycarp decided to relax on his weekend and visited to the performance of famous ropewalkers: Agafon, Boniface and Konrad.

The rope is straight and infinite in both directions. At the beginning of the performance, Agafon, Boniface and Konrad are located in positions aa , bb and cc respectively. At the end of the performance, the distance between each pair of ropewalkers was at least dd .

Ropewalkers can walk on the rope. In one second, only one ropewalker can change his position. Every ropewalker can change his position exactly by 11 (i. e. shift by 11 to the left or right direction on the rope). Agafon, Boniface and Konrad can not move at the same time (Only one of them can move at each moment). Ropewalkers can be at the same positions at the same time and can "walk past each other".

You should find the minimum duration (in seconds) of the performance. In other words, find the minimum number of seconds needed so that the distance between each pair of ropewalkers can be greater or equal to dd .

Ropewalkers can walk to negative coordinates, due to the rope is infinite to both sides.

输入格式

The only line of the input contains four integers aa , bb , cc , dd ( 1a,b,c,d1091 \le a, b, c, d \le 10^9 ). It is possible that any two (or all three) ropewalkers are in the same position at the beginning of the performance.

输出格式

Output one integer — the minimum duration (in seconds) of the performance.

输入输出样例

  • 输入#1

    5 2 6 3
    

    输出#1

    2
    
  • 输入#2

    3 1 5 6
    

    输出#2

    8
    
  • 输入#3

    8 3 3 2
    

    输出#3

    2
    
  • 输入#4

    2 3 10 4
    

    输出#4

    3
    

说明/提示

In the first example: in the first two seconds Konrad moves for 2 positions to the right (to the position 88 ), while Agafon and Boniface stay at their positions. Thus, the distance between Agafon and Boniface will be 52=3|5 - 2| = 3 , the distance between Boniface and Konrad will be 28=6|2 - 8| = 6 and the distance between Agafon and Konrad will be 58=3|5 - 8| = 3 . Therefore, all three pairwise distances will be at least d=3d=3 , so the performance could be finished within 2 seconds.

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