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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
int ctoi(const char c) {
    if ('0' <= c && c <= '9') return (c - '0');
    return -1;
}
vector<int> input(int n) {
    vector<int> vec(n);
    for (int i = 0; i < n; i++) {
        cin >> vec.at(i);
    }
    return vec;
}
void output(vector<int> vec) {
    for (int i = 0; i < vec.size(); i++) {
        cout << vec[i] << " ";
    }
    return;
}
vector<vector<int>> input(int n, int m) {
    vector<vector<int>> table(n, vector<int>(m));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> table. at(i).at(j);
        }
    }
    return table;
}
void output(vector<vector<int>> table) {
    for (int i = 0; i < table.size(); i++) {
        for (int j = 0; j < table.at(0).size(); j++) {
            cout << table.at(i).at(j) << " ";
        }
        cout << endl;
    }
}
long long perm(int n, int r) {
    if (n < r) {
        cout << "error" << endl;
        return 0;
    }
    long long perm = 1;
    for (int i = n; i > n - r; i--) {
        perm *= i;
    }
    return perm;
}
long long comb(int n, int r) {
    if (n < r) {
        cout << "error" << endl;
        return 0;
    }
    long long comb = perm(n,r);
    for (int i = r; i > 0; i--) {
        comb /= i;
    }
    return comb;
}
long long homo(int n, int r) {
    return comb(n + r - 1, n - 1);
}
long long fact(int n) {
    long long fact = 1;
    for (int i = n; i > 0; i--) {
        fact *= i;
    }
    return fact;
}
int gcd(int a, int b)
{
    if (a % b == 0)
    {
        return(b);
    }
    else
    {
        return(gcd(b, a % b));
    }
}
int lcm(int a, int b) {
    return a * b / gcd(a, b);
}
int main() {
  int n,m,x,cnt1=0,cnt2=0;
  cin >> n >> m >> x;
  vector<int> a=input(m);
  for(int i=0;i<n;i++){
    if(x<a[i]){
      cnt1++;
    }else{
      cnt2++;
    }
  }
  cout << min(cnt1,cnt2) << endl;
  
}