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)

/* << IN THE NAME OF GOD >> */
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define pob pop_back
#define ft first
#define ftt ft.ft
#define ftc ft.sc
#define sc second
#define err(x) cerr << #x  << " = "<<  x << '\n'
#define out(x) cout << x << '\n'
#define forp(i, r, l) for(int i = int(r); i < int(l); i++)
#define forn(i, r, l) for(int i = int(r); i > int(l); i--)
#define FAST ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
typedef long long ll;
#define ctz __builtin_ctz
typedef pair<int, int> pii;
typedef pair<ll ,ll> pll;
const int maxn = 1e6 + 120;
 


int n,res ,m ,x ,a[maxn];
int main(){FAST;
    cin >> n >> m >> x;
    forp(i ,0 ,m)
        cin >> a[i];
    sort(a ,a + m);
    
    forp(i ,0 ,m){
        if(a[i] > x)
        {
          break;}
        res++;
    }
    cout << min(res ,m - res);
   return 0;
}