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 "bits/stdc++.h"
using namespace std;
#define print(x) cout<<x<<endl;
#define rep(i,a,b) for(int i=a;i<b;i++)
#define REP(i,a) for(int i=0;i<a;i++)
#define debug int debug = 0;
#define RREP(i,a) for(int i=a-1;i>=0;i--)
#define rrep(i,a,b) for(int i=a-1;i>=b;i--)
#define printall(n,array) {for(int i=0;i<n;i++){cout<<array[i]<<" ";}cout<<endl;}
#define U() cout<<endl;
typedef long long ll;
typedef pair<int, int> PI;
typedef pair<int, PI> V;
typedef vector<int> VE;
const ll mod = 1000000007; //10^9+7

int main() {
    int n, m, x;
    int a[102];
    int max_ = 0, min_ = 0;
    cin >> n >> m >> x;
    REP(i, m){
        cin >> a[i];
    }
    REP(i, m){
        if(x > a[i]){
            max_++;
        }else{
            min_++;
        }
    }
    print(min(max_, min_));
}