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>

int main(int argc, char* argv[])
{
	int data[5];
	int k;

 	int ok = 1;

 	for (int n = 0; n < 5; n++) {
		scanf("%d", &data[n]);
      	if (data[n] < 0 || data[n] > 123) {
          ok *= 0;
        }
		//printf("%d\n", data[n]);
	}
	scanf("%d", &k);
	if (k < 0 || k > 123) {
		ok *= 0;
	}
	//printf("%d\n", k);

  
  	for (int i = 0; i < 4; i++) {
      for (int j = i + 1; j < 5; j++) {
        int d = data[j] - data[i];
        if (d > k) {
          ok *= 0;
        }
      }
   }
  
  	if (ok == 1) { printf("Yah!\n"); }
  	else { printf(":(\n"); }
  
	return 0;
}