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>
using namespace std;
using ll = long long;

#define rep(i,s,n) for(ll i = (s); i < (n); i++)
#define rep0(i,n) rep(i,0,n)
#define rep1(i,n) rep(i,1,n+1)
#define repR(i,s,n) for(ll i = (n-1); i >= (s); i--)
#define repR0(i,n) repR(i,0,n)
#define repR1(i,n) repR(i,1,n+1)

#define BR "\n"
#define SP " "
#define SHOW(x) for(int i = 0; i < x.size(); i++) { cout << x[i] << SP; } cout << BR;
#define SHOW2(x) for(int j = 0; j < x.size(); j++) { SHOW(x[j]); } cout << BR;
#define fcout cout << fixed << setprecision(18)

int main() {
  int N, M, X;
  cin >> N >> M >> X;

  int cnt = 0;
  rep0(i, M) {
    int a; cin >> a;
    if (a < X) cnt++;
  }
  
  int ans = min(cnt, M - cnt);
  cout << ans << BR;
  return 0;
}