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 <cstdio>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <set>
#include <string>
#include <queue>
#include <map>
#include <cmath>
#include <stack>

#define REP(i, n) for(int i = 0; i < (int)(n); i++)

using namespace std;

int main(){
  int n, m, x;
  int a[105] = {0};
  int t[105] = {0};

  cin >> n >> m >> x;
  for(int i = 0; i < m; i++){
    int work;
    cin >> work;
    a[work] = 1;
  }

  int up = x; // ひたすら上げていく
  int dn = x; // ひたすら下げていく
  while(true){
    if(up == n){
      cout << t[up] << endl;
      break;
    }else if(dn == 0){
      cout << t[0] << endl;
      break;
    }
    up++;
    t[up] = t[up - 1] + a[up];
    // cout << "UP:" << up << " " << t[up] << endl;
    dn--;
    t[dn] = t[dn + 1] + a[dn];
    // cout << "DOWN:" << dn << " " << t[dn] << endl;
  }

  return 0;
}