codeforces Round #354 (Div. 2) A

题面：

A. Nicholas and Permutation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Nicholas has an array a that contains n distinct integers from 1 to n. In other words, Nicholas has a permutation of size n.

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 integer n (2 ≤ n ≤ 100) — the size of the permutation.

The second line of the input contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ n), where ai is equal to the element at the i-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

5
4 5 1 3 2

output

3

input

7
1 6 5 3 4 7 2

output

6

input

6
6 5 4 3 2 1

output

5

Note

In the first sample, one may obtain the optimal answer by swapping elements 1 and 2.

In the second sample, the minimum and the maximum elements will be located in the opposite ends of the array if we swap 7 and 2.

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 swap 5 and 2.

题很水 就是求更换一次位置后最大最小数的距离 

发现了#include<bits/stdc++.h>原来可以用...

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define N 205
using namespace std;
int n,a1,a2,m1,m2,lm[N];
int main()
{
    scanf("%d",&n);
    a1=INF,a2=-INF;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&lm[i]);
        if(a[i]<a1) 
        {
            a1=lm[i];
            m1=i;
        }
        if(a[i]>a2) 
        {
            a2=lm[i];
            m2=i;
        }
    }
    if(m1>m2) 
        swap(m1,m2);
    int ans=max(n-m1,m2-1);
    printf("%d
",ans);
}