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>
#include <string>
#include <algorithm>
using namespace std;

typedef long long ll;
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<(n);++i)
#define REPR(i,n) for(int i=n;i>=0;i--)
#define FOREACH(x,a) for(auto& (x) : (a) )
#define ALL(obj) (obj).begin(),(obj).end()

#define P pair<int,int>
#define V vector<int>
#define M map<int,int>
#define S set<int>
#define PQ priority_queue<int>
#define PQG priority_queue<int,V,greater<int>>

#define pb(a) push_back(a)
#define mp make_pair

#define dump(x)  cout << #x << " = " << (x) << "\n";
#define COUT(x) cout<<(x)<<"\n"

int main()
{
    int n, m, x = 0;
    cin >> n >> m >> x;
    set<int> st;
    int ct1 = 0;
    int ct2 = 0;
    REP(i, m)
    {
        int buf = 0;
        cin >> buf;
        st.insert(buf);
    }
    for (int i = x; i < n; i++)
    {
        if (st.count(i) > 0)
        {
            ct1++;
        }
    }
    for (int i = x; i > 0; i--)
    {
        if (st.count(i) > 0)
        {
            ct2++;
        }

    }

    COUT(min(ct1,ct2));

    return 0;
}