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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef string str;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector <pii> vpii;
typedef vector <pll> vpll;
typedef map <str,int> mapsi;
typedef map <str,int> :: iterator mapsitr;
typedef map <int , int> mint;
typedef map <ll , ll> mll;
typedef set <int> si;
typedef set <ll> sll;
typedef si :: iterator sitr;
typedef si :: reverse_iterator rsitr;
typedef sll :: iterator sltr;
typedef sll :: reverse_iterator rsltr;
#define mset multiset
typedef mset <int> msi;
typedef mset <ll> msll;
typedef msi :: iterator msitr;
typedef msi :: reverse_iterator msritr;
typedef msll :: iterator msltr;
typedef msll :: reverse_iterator mslritr;
#define mp make_pair
#define pb push_back
#define pob pop_back
#define pf push_front
#define pof pop_front
#define fi first
#define se second
#define fs first.second
#define ss second.second
#define ff first.first
#define sf second.first
#define newl '\n'
char to_upper (char x){
    if( 97 <= int(x) && int(x) <= 126)return char(x-32);
    else if( 65 <= int(x) && int(x) <= 90)return x;
}
char to_lower (char x){
    if( 97 <= int(x) && int(x) <= 126)return x;
    else if( 65 <= int(x) && int(x) <= 90)return char(x+32);
}
int numerize (char x){
    if(48 <= int(x) && int(x) <= 57)return int(x-'0');
    else if( 97 <= int(x) && int(x) <= 126)return int(x-96);
    else if( 65 <= int(x) && int(x) <= 90)return int(x-64);
}
//void IO(){
//	#ifndef ONLINE_JUDGE
//	    freopen("input.txt", "r", stdin);
//	    freopen("output.txt", "w", stdout);
//	#endif
//}
// end of Template

int n,m,x,a,b,ans;
bool pay[200];
int main(){
	//IO();
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin>>n>>m>>x;
    for(int i=1;i<=m;i++){
        int temp;cin>>temp;
        pay[temp]=true;
    }
    for(int i=x;i>=0;i--){
        if(pay[i])a++;
    }
    for(int i=x;i<=n+1;i++){
        if(pay[i])b++;
    }
    ans = min(a,b);
    cout<<ans<<newl;

}