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.
import java.util.*;
import java.io.*;

public class Solution{
	public Solution(){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		
		int minIndex = -1;
		int maxIndex = -1;
		
		int num = -1;
		for(int i=0; i<n; i++){
			num = sc.nextInt();
			if(num == 1){
				minIndex = i;
			}
			else if(num == n){
				maxIndex = i;
			}
		}
		
		int left = maxIndex > minIndex ? maxIndex : minIndex;
		int right = maxIndex > minIndex ? minIndex : maxIndex;
		right = n - right - 1;
		
		int res = left > right ? left : right;
		
		System.out.println(res);
	}
	
	int binarySearch(int arr[], int l, int r, int x) 
    { 
        if (r >= l) { 
            int mid = l + (r - l) / 2; 
  
            // If the element is present at the 
            // middle itself 
            if (arr[mid] == x) 
                return mid; 
  
            // If element is smaller than mid, then 
            // it can only be present in left subarray 
            if (arr[mid] > x) 
                return binarySearch(arr, l, mid - 1, x); 
  
            // Else the element can only be present 
            // in right subarray 
            return binarySearch(arr, mid + 1, r, x); 
        } 
  
        // We reach here when element is not present 
        // in array 
        return -1; 
    }
	
	int partition(int arr[], int[] dep, int low, int high) 
    { 
        int pivot = arr[high];  
        int i = (low-1); // index of smaller element 
        for (int j=low; j<high; j++) 
        { 
            // If current element is smaller than or 
            // equal to pivot 
            if (arr[j] <= pivot) 
            { 
                i++; 
  
                // swap arr[i] and arr[j] 
                int temp = arr[i]; 
                arr[i] = arr[j]; 
                arr[j] = temp;
				
				if(dep != null){
					temp = dep[i];
					dep[i] = dep[j];
					dep[j] = temp;
				}
            } 
        } 
  
        // swap arr[i+1] and arr[high] (or pivot) 
        int temp = arr[i+1]; 
        arr[i+1] = arr[high]; 
        arr[high] = temp;
		
		if(dep != null){
			temp = dep[i+1];
			dep[i+1] = dep[high];
			dep[high] = temp;
		}
  
        return i+1; 
    } 
  
  
    /* The main function that implements QuickSort() 
      arr[] --> Array to be sorted, 
      low  --> Starting index, 
      high  --> Ending index */
    void sort(int arr[], int[] dep, int low, int high) 
    { 
        if (low < high) 
        { 
            /* pi is partitioning index, arr[pi] is  
              now at right place */
            int pi = partition(arr, dep, low, high); 
  
            // Recursively sort elements before 
            // partition and after partition 
            sort(arr, dep, low, pi-1); 
            sort(arr, dep, pi+1, high); 
        } 
    }
	
	public static void main(String[] args){
		new Solution();
	}
}