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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::*;
use std::str::*;

fn read<T: FromStr>(s: &mut StdinLock) -> Option<T> {
    let s = s.by_ref()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    s.parse::<T>().ok()
}

pub fn main() {
    let stdin = stdin();
    let mut stdin = stdin.lock();

    let nmx: Vec<u32> = (0..3).map(|_| (read(&mut stdin).unwrap())).collect();
    let a: Vec<u32> = (0..nmx[1]).map(|_| (read(&mut stdin).unwrap())).collect();

    let res1: u32 = a.iter().filter(|&x| x < &nmx[2]).fold(0, |acc, _| acc + 1);
    let res2: u32 = a.iter().filter(|&x| x > &nmx[2]).fold(0, |acc, _| acc + 1);

    println!("{:?}", if res1 < res2 { res1 } else { res2 });
}