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::FromStr;
use std::cmp::{min, max};
use std::mem::swap;
use std::collections::HashMap;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin_lock = stdin.lock();
    let s = stdin_lock
        .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().unwrap()
}

static DX: &'static [i32] = &[0, 0, 1, -1];
static DY: &'static [i32] = &[1, -1, 0, 0];

fn main() {
    let n: i32 = read();
    let m: i32 = read();
    let x: usize = read();

    let a: Vec<usize> = (0..m).map(|_| read()).collect();

    let left_cnt = a.iter().filter(|i| **i < x).count();
    let right_cnt = a.iter().filter(|i| x < **i).count();

    println!("{}", min(left_cnt, right_cnt));
}

fn pow_mod(mut n: u64, mut exp: u64, m: u64) -> u64 {
    let mut t = 1;

    n %= m;

    while exp > 1 {
        if exp & 1 == 1 {
            t *= n;
            t %= m;
        }

        n *= n;
        n %= m;
        exp >>= 1;
    }

    (n * t) % m
}