Problem:
Score : 200 points

Problem Statement

There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right.

Initially, you are in Square X.
You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N.
However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1.
It is guaranteed that there is no toll gate in Square 0, Square X and Square N.

Find the minimum cost incurred before reaching the goal.

Constraints

1 \leq N \leq 100

1 \leq M \leq 100

1 \leq X \leq N - 1

1 \leq A_1 < A_2 < ... < A_M \leq N

A_i \neq X

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M X
A_1 A_2 ... A_M

Output

Print the minimum cost incurred before reaching the goal.

Sample Input 1

5 3 3
1 2 4

Sample Output 1

1

The optimal solution is as follows:

First, travel from Square 3 to Square 4. Here, there is a toll gate in Square 4, so the cost of 1 is incurred.

Then, travel from Square 4 to Square 5. This time, no cost is incurred.

Now, we are in Square 5 and we have reached the goal.

In this case, the total cost incurred is 1.

Sample Input 2

7 3 2
4 5 6

Sample Output 2

0

We may be able to reach the goal at no cost.

Sample Input 3

10 7 5
1 2 3 4 6 8 9

Sample Output 3

3

Solution in Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import java.util.*;
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		//全体の長さを入力
		int n = sc.nextInt();
		
		//料金所の数を設定
		int m = sc.nextInt();
		
		//初期位置を設定
		int x = sc.nextInt();
		
		//料金所の位置を設定
		int a[] = new int[m];
		for(int i = 0;i < m;i++) {
			a[i] = sc.nextInt();
		}
		
		//料金所通過のカウント
		int fee = 0;
		
		//前に進むか後ろに進むかの判断
		boolean forward = false;
		boolean back = false;
		if(x < n-x) {
			back = true;
		}else {
			forward = true;
		}
		
		//実際に動かす
		//前の時
		if(forward) {
			while(true) {
				x++;
				if(x == n) {
					break;
				}
				for(int i = 0;i < m;i++) {
					if(x == a[i]) {
						fee++;
					}
				}
			}
		}
		//後ろの時
		if(back) {
			while(true) {
				x--;
				if(x == 0) {
					break;
				}
				for(int i = 0;i < m;i++) {
					if(x == a[i]) {
						fee++;
					}
				}
			}
		}
		System.out.println(fee);
		sc.close();
	}
}