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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

typedef long long ll;
typedef int keytype;

int gcd(int a, int b){ return (a % b) ? gcd(b, a % b) : b;}
int lcm(int a, int b){ return (a / gcd(a, b)) * b;}

/* a[0], ..., a[n-1] の数の最大公約数 */
int ngcd(int n, int a[])
{
    int i, d;

    d = a[0];
    for (i = 1; i < n && d != 1; i++)
        d = gcd(a[i], d);
    return d;
}

void quicksort(keytype a[], int first, int last)
{
    int i, j;
    keytype x, t;

    x = a[(first + last) / 2];
    i = first;  j = last;
    for ( ; ; ) {
        while (a[i] < x) i++;
        while (x < a[j]) j--;
        if (i >= j) break;
        t = a[i];  a[i] = a[j];  a[j] = t;
        i++;  j--;
    }
    if (first  < i - 1) quicksort(a, first , i - 1);
    if (j + 1 < last) quicksort(a, j + 1, last);
}

int is_prime(int x){
    if(x <= 1) return 0;
    for(int i = 2; i * i <= x; i++){
        if(x % i == 0) return 0;
    }
    return 1;
}

int is_substr(char s[], char t[], int i){
    int flag = 1;

    if(s[i] == '\0') flag = 0;

    for(int j = 0; j < strlen(t); ++j, ++i){
        if(s[i] != t[j]){
            flag = 0;
            break;
        }
    }
    return flag;
}

// factor list, number of factors, natural number
void factorize(int ftr[], int *num, int n){
    int a = 2, cnt = 0;

    while(n >= a){
        if(n % a == 0){
            n = n / a;
            ftr[cnt++] = a;
        }else{
            a++;
        }
    }
    *num = cnt;
}

//#define N 5
//int p[N + 1];
int nextperm(int p[], int N)  /* 辞書式順序で次の順列 */
{
    int i, j, t;

    i = N - 1;
    p[0] = 0;  /* 番人 */
    while (p[i] >= p[i + 1]) i--;
    if (i == 0) return 0;  /* 完了 */
    j = N;
    while (p[i] >= p[j]) j--;
    t = p[i];  p[i] = p[j];  p[j] = t;
    i++;  j = N;
    while (i < j) {
        t = p[i];  p[i] = p[j];  p[j] = t;  i++;  j--;
    }
    return 1;  /* 未了 */
}

void comb(int size1, int size2, int v[size1][size2]){
  for(int i = 0; i < size1; i++){
    v[i][0] = 1;
    v[i][i] = 1;
  }
  for(int k = 1; k < size1; k++){
    for(int j = 1; j < k; j++){
      v[k][j] = (v[k-1][j-1] + v[k-1][j]);
    }
  }
}

int abs_val(int a, int b){
    if(a > b) return a - b;
    else return b - a;
}

const int inf = 1012345678;
int main(){
    int p[5];
    int index[6] = {0, 1, 2, 3, 4, 5};
    int k;

    scanf("%d %d %d %d %d %d", &p[0], &p[1], &p[2], &p[3], &p[4], &k);

    for(int i = 0; i < 5; ++i){
        for(int j = i + 1; j < 5; ++j){
            if((p[j] - p[i]) > k){
                puts(":(");
                return 0;
            }
        }
    }

    puts("Yay!");

    return 0;
}