codeforces 676A Nicholas and Permutation

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 containsn distinct integers from1 to n. 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

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 elements1 and 2.

In the second sample, the minimum and the maximum elements will be located in the opposite ends of the array if we swap7 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 swap5 and 2.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <map>
#include <set>

using namespace std;

int a[100];

int main()
{
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}
	int maxn = -1, minn = 0x7fffffff, ia = 0, in = 0;
	for (int i = 0; i < n; i++) {
		if (a[i] > maxn) {
			maxn = a[i];
			ia = i;
		}
		if (a[i] < minn) {
			minn = a[i];
			in = i;
		}
	}
	if (ia == 0 || ia == n - 1 || in == 0 || in == n - 1) {
		cout << n - 1 << endl;
	} else {
		cout << max(max(in, ia), max(n - 1 - in, n - 1 - ia)) << endl;
	}
	return 0;
}