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>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = s; i < (int)(n); i++)
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()

using namespace std;
using ll = long long;
using pii = pair<int, int>;

//const ll INF = 1e18L + 5;
//const int INF = 1e9 + 5;
//const double pi = 3.14159265358979323846;

int		main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);

	int n, m, x;
	cin >> n >> m >> x;
	int A = 0, B = 0;
	rep(i, m) {
		int a;
		cin >> a;
		if (a < x)
			A++;
		else
			B++;
	}
	cout << min(A, B) << endl;
}