Nicholas and Permutation

                                                                                  

                                                                                                    A. Nicholas and Permutation

Nicholas has an array a that containsndistinct integers from1 ton. In other words, Nicholas has a permutation of sizen.

Nicholas want the minimum element (integer 1) and the maximum element (integer n) to be as far as possible from each other. He wants to perform exactly one swap in order to maximize the distance between the minimum and the maximum elements. The distance between two elements is considered to be equal to the absolute difference between their positions.

Input

The first line of the input contains a single integern (2 ≤ n ≤ 100) — the size of the permutation.

The second line of the input contains n distinct integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ n), where a_i is equal to the element at thei-th position.

Output

Print a single integer — the maximum possible distance between the minimum and the maximum elements Nicholas can achieve by performing exactly one swap.

Examples

Input

54 5 1 3 2

Output

3

Input

71 6 5 3 4 7 2

Output

6

Input

66 5 4 3 2 1

Output

5

Note

In the first sample, one may obtain the optimal answer by swapping elements1 and2.

In the second sample, the minimum and the maximum elements will be located in the opposite ends of the array if we swap7 and2.

In the third sample, the distance between the minimum and the maximum elements is already maximum possible, so we just perform some unnecessary swap, for example, one can swap5 and2.

题意：

交换这个序列里面的两个数使最大的数和最小的数的距离最大,问这个最大的距离是多少;

思路：

把最大的和最小的分别和第一个和最后一个交换,最大的就是答案了;

ＡＣ代码：

#include<iostream>#include<string.h>#include<cmath>#include<stdio.h>#include<algorithm>using namespace std;int main(){int a[101];int T;//int ma,mi,mid,dis;int Max,Min;while(cin>>T){memset(a,0,sizeof(a));for(int i=1;i<=T;i++){cin>>a[i];if(a[i]==1)Min=i;if(a[i]==T)Max=i;}//cout<<mi<<ma<<endl;if((Max-Min)==T-1)cout<<T-1<<endl;else{int ans=max(T-Min,T-Max);ans=max(ans,Min-1);ans=max(ans,Max-1);cout<<ans<<endl;}}}