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<iostream>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<vector>
#include<list>
#include<functional>
#include<algorithm>
#include<string>
#include<cmath>
#include<complex>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<deque>
#include<tuple>
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n) for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int, int> P;



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

	int N, M, X;
	cin >> N >> M >> X;
	vector<int> memo(N + 1, 0);
	REP(i, M) {
		int A;
		cin >> A;
		memo[A]++;
	}
	int cost0 = 0, costN = 0;
	REP(i, X) cost0 += memo[i];
	FOR(i, X + 1, N + 1) costN += memo[i];
	cout << min(cost0, costN) << "\n";

	return 0;
}