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;

// B - Toll Gates
// https://atcoder.jp/contests/abc094/tasks/abc094_b
public class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int N = Integer.parseInt(scanner.next());
        int M = Integer.parseInt(scanner.next());
        int X = Integer.parseInt(scanner.next());
        boolean[] A = new boolean[N + 1];
        for (int i = 0; i < M; i++) {
            int index = Integer.parseInt(scanner.next());
            A[index] = true;
        }
        scanner.close();

        int costA = 0;
        for (int i = X; i <= N; i++) {
            if (A[i]) costA++;
        }
        int costB = 0;
        for (int i = X; i >= 0; i--) {
            if (A[i]) costB++;
        }

        System.out.println(Math.min(costA, costB));
    }
}