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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
typedef long long int ll;
void swap(ll *x,ll *y){ll temp;temp=*x;*x=*y;*y=temp;}
ll asc(const void *a,const void *b){return *(ll*)a-*(ll*)b;}
ll desc(const void *a,const void *b){return *(ll*)b-*(ll*)a;}
ll max(ll x,ll y){return x>y?x:y;}
ll min(ll x,ll y){return x>y?y:x;}

int main(){
    int n,m,x,a[200];
    int mass[200];

    for(int i=0;i<n+1;i++)mass[i]=0;
    scanf("%d %d %d",&n,&m,&x);
    for(int i=0;i<m;i++){
        scanf("%d",&a[i]);
        mass[a[i]-1]=1;
    }

    int sum1,sum2;
    sum1=sum2=0;
    for(int i=0;i<x;i++)sum1+=mass[i];
    for(int i=x-1;i<n;i++)sum2+=mass[i];

    if(sum1>sum2)printf("%d\n",sum2);
    else printf("%d\n",sum1);

    return 0;
}