Problem:
Score: 200 points

Problem Statement

A country decides to build a palace.

In this country, the average temperature of a point at an elevation of x meters is T-x \times 0.006 degrees Celsius.

There are N places proposed for the place. The elevation of Place i is H_i meters.

Among them, Princess Joisino orders you to select the place whose average temperature is the closest to A degrees Celsius, and build the palace there.

Print the index of the place where the palace should be built.

It is guaranteed that the solution is unique.

Constraints

1 \leq N \leq 1000

0 \leq T \leq 50

-60 \leq A \leq T

0 \leq H_i \leq 10^5

All values in input are integers.

The solution is unique.

Input

Input is given from Standard Input in the following format:

N
T A
H_1 H_2 ... H_N

Output

Print the index of the place where the palace should be built.

Sample Input 1

2
12 5
1000 2000

Sample Output 1

1

The average temperature of Place 1 is 12-1000 \times 0.006=6 degrees Celsius.

The average temperature of Place 2 is 12-2000 \times 0.006=0 degrees Celsius.

Thus, the palace should be built at Place 1.

Sample Input 2

3
21 -11
81234 94124 52141

Sample Output 2

3

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

macro_rules! input {
  (source = $s:expr, $($r:tt)*) => {
      let mut iter = $s.split_whitespace();
      let mut next = || { iter.next().unwrap() };
      input_inner!{next, $($r)*}
  };
  ($($r:tt)*) => {
      let stdin = std::io::stdin();
      let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
      let mut next = move || -> String{
          bytes
              .by_ref()
              .map(|r|r.unwrap() as char)
              .skip_while(|c|c.is_whitespace())
              .take_while(|c|!c.is_whitespace())
              .collect()
      };
      input_inner!{next, $($r)*}
  };
}

macro_rules! input_inner {
  ($next:expr) => {};
  ($next:expr, ) => {};

  ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
      let $var = read_value!($next, $t);
      input_inner!{$next $($r)*}
  };
}

macro_rules! read_value {
  ($next:expr, ( $($t:tt),* )) => {
      ( $(read_value!($next, $t)),* )
  };

  ($next:expr, [ $t:tt ; $len:expr ]) => {
      (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
  };

  ($next:expr, chars) => {
      read_value!($next, String).chars().collect::<Vec<char>>()
  };

  ($next:expr, usize1) => {
      read_value!($next, usize) - 1
  };

  ($next:expr, $t:ty) => {
      $next().parse::<$t>().expect("Parse error")
  };
}

pub trait LexicalPermutation {
  /// Return `true` if the slice was permuted, `false` if it is already
  /// at the last ordered permutation.
  fn next_permutation(&mut self) -> bool;
  /// Return `true` if the slice was permuted, `false` if it is already
  /// at the first ordered permutation.
  fn prev_permutation(&mut self) -> bool;
}

impl<T> LexicalPermutation for [T]
where
  T: Ord,
{
  /// Original author in Rust: Thomas Backman <serenity@exscape.org>
  fn next_permutation(&mut self) -> bool {
      // These cases only have 1 permutation each, so we can't do anything.
      if self.len() < 2 {
          return false;
      }

      // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
      let mut i = self.len() - 1;
      while i > 0 && self[i - 1] >= self[i] {
          i -= 1;
      }

      // If that is the entire vector, this is the last-ordered permutation.
      if i == 0 {
          return false;
      }

      // Step 2: Find the rightmost element larger than the pivot (i-1)
      let mut j = self.len() - 1;
      while j >= i && self[j] <= self[i - 1] {
          j -= 1;
      }

      // Step 3: Swap that element with the pivot
      self.swap(j, i - 1);

      // Step 4: Reverse the (previously) weakly decreasing part
      self[i..].reverse();

      true
  }

  fn prev_permutation(&mut self) -> bool {
      // These cases only have 1 permutation each, so we can't do anything.
      if self.len() < 2 {
          return false;
      }

      // Step 1: Identify the longest, rightmost weakly increasing part of the vector
      let mut i = self.len() - 1;
      while i > 0 && self[i - 1] <= self[i] {
          i -= 1;
      }

      // If that is the entire vector, this is the first-ordered permutation.
      if i == 0 {
          return false;
      }

      // Step 2: Reverse the weakly increasing part
      self[i..].reverse();

      // Step 3: Find the rightmost element equal to or bigger than the pivot (i-1)
      let mut j = self.len() - 1;
      while j >= i && self[j - 1] < self[i - 1] {
          j -= 1;
      }

      // Step 4: Swap that element with the pivot
      self.swap(i - 1, j);

      true
  }
}

#[allow(unused_imports)]
use std::cmp::{max, min};
use std::mem::swap;

fn main() {
	input!{
    	n: i32,
      	t: i32,
      	a: i32,
      	h: [f32; n],
	}	
	let mut h = h;
	let mut ans = 0;
	let mut sum = (t - (h[0] * 0.006) as i32).abs();
  
	for i in 1..h.len() {
    	let tmp = (t - (h[i] * 0.006) as i32).abs();
        if sum > tmp {
            ans = i;
            sum = tmp;
        }
    }
  	println!("{:?}", ans + 1);
}