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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
int min(int a, int b)
{
  if(a <= b)
  return a;
  else
  return b;
}

void swap(int *p_from, int *p_to) {
    int tmp;
    tmp = *p_from;
    *p_from = *p_to;
    *p_to = tmp;
}

int main(void)
{
  int a[5];
  int i;
  int min, c, minp;
  c = 0;
  minp = 5;
  for(i = 0; i < 5; i++)
  scanf("%d", &a[i]);
  min = 9;
  for(i = 0; i < 5; i++){
    if(a[i] % 10 <= min && a[i] % 10 != 0){
    min = a[i] % 10;
    minp = i;
  }
  }
  for(i = 0; i < 5; i++){
    if(a[i] % 10 != 0 && i != minp)
    c += (a[i] / 10)*10 +10;
    else
    c += a[i];
  }
  printf("%d\n", c);
  return 0;
}