Problem:
﻿Score: 100 points

Problem Statement

In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.

Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.

Determine if there exists a pair of antennas that cannot communicate directly.

Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.

Constraints

a, b, c, d, e and k are integers between 0 and 123 (inclusive).

a < b < c < d < e

Input

Input is given from Standard Input in the following format:

a
b
c
d
e
k

Output

Print :( if there exists a pair of antennas that cannot communicate directly, and print Yay! if there is no such pair.

Sample Input 1

1
2
4
8
9
15

Sample Output 1

Yay!

In this case, there is no pair of antennas that cannot communicate directly, because:

the distance between A and B is 2 - 1 = 1

the distance between A and C is 4 - 1 = 3

the distance between A and D is 8 - 1 = 7

the distance between A and E is 9 - 1 = 8

the distance between B and C is 4 - 2 = 2

the distance between B and D is 8 - 2 = 6

the distance between B and E is 9 - 2 = 7

the distance between C and D is 8 - 4 = 4

the distance between C and E is 9 - 4 = 5

the distance between D and E is 9 - 8 = 1

and none of them is greater than 15. Thus, the correct output is Yay!.

Sample Input 2

15
18
26
35
36
18

Sample Output 2

:(

In this case, the distance between antennas A and D is 35 - 15 = 20 and exceeds 18, so they cannot communicate directly.
Thus, the correct output is :(.

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

fn main() {
    let v: [i32; 5] = {
        let m = read_mat::<i32>(5);
        [
            m[0][0],
            m[1][0],
            m[2][0],
            m[3][0],
            m[4][0],
        ]
    };

    // 最後に頼むべき品を見つける
    let highest_cost_dish: usize = v.into_iter()
        .map(|x| x % 10 + 10)
        .enumerate()
        .min_by_key(|x| (x.1 - 1) % 10 )
        .unwrap()
        .0;

    let ans = v
        .into_iter()
        .enumerate()
        .map(|(i, &v)| {
            if i != highest_cost_dish {
                (i, 10 * ((v + (10 - 1)) / 10))
            } else {
                (i, v)
            }
        })
        .fold(0, |acc, x| acc + x.1);

    println!("{}", ans);
}

#[allow(dead_code)]
fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

#[allow(dead_code)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().split_whitespace()
        .map(|e| e.parse().ok().unwrap()).collect()
}

#[allow(dead_code)]
fn read_mat<T: std::str::FromStr>(row: u32) -> Vec<Vec<T>> {
    let mut vec = Vec::new();

    for _i in 0..row {
        vec.push(read_vec::<T>());
    }

    vec
}