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 C (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>

int check(int *a, int x)
{
	int i = 0;
	while (a[i])
	{
		if (x == a[i])
			return 1;
		i++;
	}
	return (0);
}

// int		main(void)
// {
// 	int n,m,x;
// 	scanf("%d%d%d", &n, &m, &x);
// 	int a[m];
// 	int i = 0;
// 	int count1 = 0;
// 	int count2 = 0;
// 	while (i < m)
// 		scanf("%d", &a[i++]);
// 	i = x;
// 	int zero = a[0] - 1;
// 	if (x < n)
// 	{
// 		while (i < x)
// 		{
// 			if (check(a, i))
// 				count1++;
// 			i++;
// 		}
// 	}
// 	else
// 	{
// 		while (i > n)
// 		{
// 			if (check(a, i))
// 				count1++;
// 			i--;
// 		}
// 	}
// 	i = x;
// 	while (i > zero)
// 	{
// 		if (check(a, i))
// 			count2++;
// 		i--;
// 	}
// 	if (count1 < count2)
// 		printf("%d\n", count1);
// 	else
// 		printf("%d\n", count2);
// 	return (0);
// }

int		main(void)
{
	int n,m,x;
	scanf("%d%d%d", &n, &m, &x);
	int a[m];
	int i = 0;
	int count1 = 0;
	int count2 = 0;
	while (i < m)
		scanf("%d", &a[i++]);
	i = 0;
	int zero = a[0] - 1;
	while (i < m)
	{
		if(x < a[i])
			count1++;
		else
			count2++;
		i++;
	}
	// i = x;
	// while (i > zero)
	// {
	// 	if (check(a, i))
	// 		count2++;
	// 	i--;
	// }
	if (count1 < count2)
		printf("%d\n", count1);
	else
		printf("%d\n", count2);
	return (0);
}