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 <iostream>
#include <cmath>
#include <string>
#include <algorithm>
#include <limits>
#include <ctype.h> 
#include <stdio.h> 
#include <set> 
#include <iomanip>
#include <vector>
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
#define L(ID) ((ID*2)+1)
#define R(ID) ((ID*2)+2)
#define M(L,R) ((L+R)/2)
#define forr(start,end) for(int i = start; i < end; i++)
#define gap ' '

const int MAXN = 4*1e5;

//ll a[MAXN], ST[MAXN * 4], LZ[MAXN*4];

/*bool perfectSquare(ll x){
    ll s = sqrt(x);
    if(s *s == x) return true;
    else return false;
}*/
typedef long double ld;

ll a[MAXN], n = 0;


ll gcd(ll a, ll b) 
{ 
    if (a == 0) 
        return b; 
    return gcd(b % a, a); 
}


int main() { 
    int n = 0, m = 0, x = 0;
    cin >> n >> m >> x;
    int a = 0, l = 0, r = 0;
    for(int i = 0; i < m; ++i){
        cin >> a;
        if(a < x){
            ++l;
        }
        else{
            ++r;
        }
    }
    cout << min(l,r) << endl;
    

    return 0; 
}