最小秒數

Luba is surfing the Internet. She currently has n opened tabs in her browser, indexed from 1 to n from left to right. The mouse cursor is currently located at the pos-th tab. Luba needs to use the tabs with indices from l to r (inclusive) for her studies, and she wants to close all the tabs that don’t belong to this segment as fast as possible.

Each second Luba can either try moving the cursor to the left or to the right (if the cursor is currently at the tab i, then she can move it to the tab max(i - 1, a) or to the tab min(i + 1, b)) or try closing all the tabs to the left or to the right of the cursor (if the cursor is currently at the tab i, she can close all the tabs with indices from segment [a, i - 1] or from segment [i + 1, b]). In the aforementioned expressions a and b denote the minimum and maximum index of an unclosed tab, respectively. For example, if there were 7 tabs initially and tabs 1, 2 and 7 are closed, then a = 3, b = 6.

What is the minimum number of seconds Luba has to spend in order to leave only the tabs with initial indices from l to r inclusive opened?

Input
The only line of input contains four integer numbers n, pos, l, r (1 ≤ n ≤ 100, 1 ≤ pos ≤ n, 1 ≤ l ≤ r ≤ n) — the number of the tabs, the cursor position and the segment which Luba needs to leave opened.

Output
Print one integer equal to the minimum number of seconds required to close all the tabs outside the segment [l, r].

Examples
Input
6 3 2 4
Output
5
Input
6 3 1 3
Output
1
Input
5 2 1 5
Output
0
Note
In the first test Luba can do the following operations: shift the mouse cursor to the tab 2, close all the tabs to the left of it, shift the mouse cursor to the tab 3, then to the tab 4, and then close all the tabs to the right of it.

In the second test she only needs to close all the tabs to the right of the current position of the cursor.

In the third test Luba doesn’t need to do anything.
思路：
讀題讀了好久。。。
當鼠標在pos位置的時候，可以有四種操作，每種操作花費1秒
1.向左移動一次
2.向右移動一次
3.關閉當前位置左邊的所有網頁（包括該網頁）
4.關閉當前位置右邊的所有網頁（包括該網頁）

#include<bits/stdc++.h>
using namespace std;
const int maxx=110;

int main()
{
   int n,pos,l,r;
   cin>>n>>pos>>l>>r;
   long long int sum=0;
   if(l==1&&r==n)
    sum=0;
   else if(l==1)//左端點在1處，只需要將鼠標移動到r+1處，並關閉r右邊的所有網頁
   {
       sum=abs(r-pos)+1;
   }
   else if(r==n)//右端點在n處，只需要將鼠標移動到l-1處，並關閉l左邊的所有網頁
   {
       sum=abs(pos-l)+1;
   }
   else//鼠標先移動到l與r中離pos最近的一個上面，然後再往左或往右移動一個，關閉所有的網頁，然後再將鼠標移動到另一個端點+1上面，關閉所有標籤
   {
       sum=min(abs(pos-l),abs(r-pos))+1+(r-l)+1;
   }
   cout<<sum<<endl;
    return 0;
}