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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <math.h>
#include <map>
#include <bitset>
#include <cmath>
#include <set>
#include <utility>

#define INF 1000000000000
#define MOD 1000000007 //10^9+7:合同式の法

using namespace std;
typedef long long ll;

//マクロ
#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define REPD(i,n) for(ll i=n-1;i>=0;i--)
#define FOR(i,a,b) for(ll i=a;i<=(ll)(b);i++)
#define FORD(i,a,b) for(ll i=a;i>=(ll)(b);i--)

int main(){
    int N,M,X,min = 10000,rs=0,t = 0;
    cin >> N >> M >> X;
    vector<int> A(M);

    for(int i = 0;i < M;i++)
        cin >> A.at(i);

    for(int i = 0;;i++){
        if(A.at(i) < X)
            rs++;
        else
        {
            t = i;
            break;
        }       
    }

    if(min > rs)
        min = rs;
    rs = 0;

    for(int i = t;i < A.size();i++){
        if(A.at(i) < N)
            rs++;
    }

    if(min > rs){
        min = rs;
    }

    cout << min << endl;

    

}