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>

void Transport(double,double*);

int main(){
  double counter = 0;
  double city_num[] = {0,0,0,0,0,0};
  double* num;
  double N,A,B,C,D,E;

  scanf("%lf",&N);
  scanf("%lf",&A);
  scanf("%lf",&B);
  scanf("%lf",&C);
  scanf("%lf",&D);
  scanf("%lf",&E);

  city_num[0] = N;
  num = city_num;

  while(1){
    counter ++;
    Transport(E,num + 4);
    Transport(D,num + 3);
    Transport(C,num + 2);
    Transport(B,num + 1);
    Transport(A,num);

    if(city_num[5] == N){
      break;
    }
  }

  printf("%f",counter);

  return 0;
}

void Transport(double X, double* x){
  double keep;
  if(*x < X){
    keep = *x;
    *x = 0;
    *(x + 1) += keep;
  }else{
    *x -= X;
    *(x + 1) += X;
  }
}