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 <algorithm>
#include <iostream>
#include <map>
#include <vector>
#define REP(i, n) for (int i = 0; i < (n); i++)
using namespace std;

void solve(long long N, long long M, long long X, std::vector<long long> A){
    map<int, int> m;
    REP(i, N + 1)
        m.insert(make_pair(i, 0));
    REP(i, M)
        m[A[i]] = 1;

    long long ans1 = 0;
    REP(i, X)
        ans1 += m[i];

    long long ans2 = 0;
    for (int i = X; i < N; i++)
        ans2 += m[i];

    cout << min(ans1, ans2) << '\n';
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    long long N;
    scanf("%lld",&N);
    long long M;
    scanf("%lld",&M);
    long long X;
    scanf("%lld",&X);
    std::vector<long long> A(M);
    for(int i = 0 ; i < M ; i++){
        scanf("%lld",&A[i]);
    }
    solve(N, M, X, std::move(A));
    return 0;
}