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)

/*
Author : N_o_o_B
Created : April 22 2020 16:04:57 
*/
#include <bits/stdc++.h>
using namespace std;
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
*/
#define TRACE
#ifdef TRACE
#define trace(...) { cerr<<"[ "; __trace__(#__VA_ARGS__, __VA_ARGS__);}
#undef endl
template <typename Arg1,typename Arg2>
ostream& operator << (ostream& out, const pair<Arg1,Arg2> &x) {
    return out<<"("<<x.first<<","<<x.second<<")";
}
 
template <typename Arg1>
ostream& operator << (ostream& out, const vector<Arg1> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1>
ostream& operator << (ostream& out, const set<Arg1> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1,typename Arg2>
ostream& operator << (ostream& out, const map<Arg1,Arg2> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1>
void __trace__(const string name, Arg1&& arg1){
    cerr << name << " : " << arg1 << " ] " << endl;
}
 
template <typename Arg1, typename... Args>
void __trace__(const string names, Arg1&& arg1, Args&&... args){
    const string name = names.substr(0,names.find(','));
    cerr<<name<<" : "<<arg1<<" | ";
    __trace__(names.substr(1+(int)name.size()), args...);
}
#else
    #define trace(args...)
#endif
 
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<pair<int,int>> vii;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<pair<ll,ll>> vll;
typedef vector<vl> vvl;
//typedef tree<pii, null_type, less<pii>, rb_tree_tag, tree_order_statistics_node_update> oset;
#define fori(i,n) for(int i=0;i<n;i++)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define ford(i,n)  for(int i=n-1;i>=0;i--)
#define trav(x,a) for(auto& x:a)
#define all(x) x.begin(),x.end()
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define fi first
#define se second
clock_t time_p=clock();
void time_taken()
{
    time_p=clock()-time_p;
    cerr<<"Time Taken : "<<(float)(time_p)/CLOCKS_PER_SEC<<"\n";
}
//const ll mod=998244353;
const ll mod=1e9+7;
const ll INF=1e18;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);cout.tie(NULL);
    cout.precision(12);cout << fixed;

    int n,m,x;
    cin >> n >> m >> x;
    vi a(n+1);
    fori(i,m){
        int x;
        cin >> x;
        a[x]=1;
    }
    int ans1 = 0;
    fori(i,x){
        ans1 += a[i];
    }
    cout << min(ans1,m-ans1) << endl;

    time_taken();
    return 0;
}