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

public class Task676A {

    public static void main(String... args) throws NumberFormatException,
            IOException {
        Solution.main(System.in, System.out);
    }

    static class Strings {
        static boolean isPalindrome(String inp) {
            for (int i = 0; i < inp.length(); i++) {
                if (inp.charAt(i) != inp.charAt(inp.length() - 1 - i)) {
                    return false;
                }
            }
            return true;
        }
    }

    static class Scanner implements AutoCloseable {

        private final BufferedReader br;
        private String[] cache;
        private int cacheIndex;

        Scanner(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
            cache = new String[0];
            cacheIndex = 0;
        }

        int nextInt() throws IOException {
            if (cacheIndex >= cache.length) {
                cache = br.readLine().split(" ");
                cacheIndex = 0;
            }
            return Integer.parseInt(cache[cacheIndex++]);
        }

        long nextLong() throws IOException {
            if (cacheIndex >= cache.length) {
                cache = br.readLine().split(" ");
                cacheIndex = 0;
            }
            return Long.parseLong(cache[cacheIndex++]);
        }

        String next() throws IOException {
            if (cacheIndex >= cache.length) {
                cache = br.readLine().split(" ");
                cacheIndex = 0;
            }
            return cache[cacheIndex++];
        }

        public void close() throws IOException {
            br.close();
        }

    }

    static class Solution {


        public static int GCD(int a, int b) {
            return b == 0 ? a : GCD(b, a % b);
        }

        public static void main(InputStream is, OutputStream os)
                throws NumberFormatException, IOException {

            try (PrintWriter pw = new PrintWriter(os)) {
                try (Scanner sc = new Scanner(is)) {


                    int n = sc.nextInt();
                    int pos1 = 0;
                    int posn = 0;
                    for (int i = 1; i <= n; i++) {
                        int k = sc.nextInt();
                        if (k == 1) {
                            pos1 = i;
                        } else if (k == n) {
                            posn = i;
                        }
                    }
                    pw.println(Math.max(Math.max(n - posn, posn - 1), Math.max(n - pos1, pos1 - 1)));
                }
            }
        }
    }
}