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: Wrong Answer)

#pragma region header
#define _GLIBCXX_DEBUG 
#include <bitset>
#include <tuple>
#include <cstdint>
#include <cstdio>
#include <cctype>
#include <assert.h>
#include <stdlib.h>
#include <stdio.h>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <ctime>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <limits>
#include <map>
#include <memory>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#include <math.h>
using namespace std;
#define int long long
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define rev(i, n) for(int i = (int)(n - 1); i >= 0; i--)
#define rev1(i, n) for(int i = (int)(n); i > 0; i--)
#define pb push_back
#define ts to_string
#define all(v) (v).begin(), (v).end()
#define resort(v) sort((v).rbegin(), (v).rend())
#define vi vector<int>
#define vvi vector<vector<int>>
#define vc vector<char>
#define vvc vector<vector<char>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define vs vector <string>
#define vvs vector <vector<string>>
#define pq  priority_queue<int>
#define smallpq  priority_queue<int, vector<int>, greater<int>>
using ll = long long;
using P = pair<int, int>;
/* ----------------よく使う数字や配列----------------- */
int dx[] = { 1,0,-1,0 };
int dy[] = { 0,1,0,-1 };
constexpr ll mod = 1e9 + 7;
constexpr ll inf = INT32_MAX / 2;
constexpr ll INF = LLONG_MAX / 2;
constexpr long double eps = DBL_EPSILON;
constexpr long double pi = 3.141592653589793238462643383279;
/* ----------------------end----------------------- */
/* --------------------テンプレート------------------ */
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
/* ----------------------end----------------------- */

/* --------------------ライブラリ-------------------- */
ll fact(int i) {       //階乗
    if (i == 0) return 1;
    return (fact(i - 1)) * i % mod;
}
ll gcd(ll a, ll b) {        //最大公約数
    if (b == 0) return a;
    return gcd(b, a % b);
}
ll lcm(ll a, ll b) {      //最小公倍数
    return a * b / gcd(a, b);
}
bool isprime(ll a)//素数判定、primeならtrue,違うならfalse
{
    if (a < 2) return false;
    else if (a == 2) return true;
    else if (a % 2 == 0) return false;
    double m = sqrt(a);
    for (int i = 3; i <= m; i += 2)
    {
        if (a % i == 0)
        {
            return false;
        }
    }
    // 素数である
    return true;
}
ll keta(ll n) {        //桁数を求める
    if (n == 0) return 1;
    int count = 0;
    while (n != 0) {
        n /= 10;
        count++;
    }
    return count;
}
ll ketasum(ll n) {    //各桁の和
    ll sum = 0;
    while (n != 0) {
        sum += n % 10;
        n /= 10;
    }
    return sum;
}
/* ----------------------end----------------------- */
#pragma endregion
signed main() {
    vi H(5);
    int k;
    rep(i, 5) {
        cin >> H[i];
    }
    cin >> k;
    bool b = true;
    rep(i, 4) {
        for (int j = 0; j > i; j++) {
            if (H[j] - H[i] > k) b = false;
        }
    }
    if (b) puts("Yay!");
    else puts(":(");
}