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){
		//N＋1個の箱
		//n+1個の左からの宛名0,1,....,Nとなる
		//現在位置X左右自由に移動可
		//宛名0とN(0<N<100)で終了
		//関所Ai(i = 1,2,.....,M(0<M<100)) == aの時1コスト
		//Ai と 0,X,Nは==でない
		//
		//nもしくは0に近い方にいく(0+N)/2>Amの時,
		//(0+N)/2<Amの時でxの位置確認
		//１、まずXの位置だす

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt()+1;//箱
		int m = sc.nextInt();//関所の住所
		int x = sc.nextInt()-1;//Xの位置と配列の誤差を消す
		int evalu[] = new int[m];
		int a[] = new int[m];
		int light = 0;
		int left = 0;
		for(int i = 0;i < m;i++){
			a[i] = sc.nextInt();//A1からAmゲット
		//１、まずXの位置だす
			if(a[i]>x){//Xの位置より右にいくつ料金所があるか算出

				light++;
			}else if(a[i]<x){//Xの位置より左にいくつ料金所があるか算出
				left++;
			}
		}
		//１、まずXの位置だす
		if (light == left){
			System.out.println(left);	
		}else if(light > left){
			System.out.println(left);
		}else if(light < left){
			System.out.println(light);
		}
	}
}