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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;
import java.util.Arrays;

public class Main {

    public static void main(String[] args) {
        new Main().solve();
    }

    void solve() {
        Scanner sc = new Scanner(System.in);
        int N=sc.nextInt();
        boolean[] A=new boolean[N];
        Arrays.fill(A, false);
        int M=sc.nextInt();
        int X=sc.nextInt();
        X--;
        for(int i=0;i<M;i++){
            int a=sc.nextInt();
            a--;
            A[a]=true;
        }
        int ans=0;
        for(int i=X;i<N;i++){
            if(A[i]==true)ans++;
        }
        int answer=0;
        for(int i=X;i>=0;i--){
            if(A[i]==true)answer++;
        }
        System.out.println(Math.min(ans, answer));
    }
}