最小秒数

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：
读题读了好久。。。
当鼠标在pos位置的时候，可以有四种操作，每种操作花费1秒
1.向左移动一次
2.向右移动一次
3.关闭当前位置左边的所有网页（包括该网页）
4.关闭当前位置右边的所有网页（包括该网页）

#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)//左端点在1处，只需要将鼠标移动到r+1处，并关闭r右边的所有网页
   {
       sum=abs(r-pos)+1;
   }
   else if(r==n)//右端点在n处，只需要将鼠标移动到l-1处，并关闭l左边的所有网页
   {
       sum=abs(pos-l)+1;
   }
   else//鼠标先移动到l与r中离pos最近的一个上面，然后再往左或往右移动一个，关闭所有的网页，然后再将鼠标移动到另一个端点+1上面，关闭所有标签
   {
       sum=min(abs(pos-l),abs(r-pos))+1+(r-l)+1;
   }
   cout<<sum<<endl;
    return 0;
}