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	`1`	`+The maximum distance between any two antenna is A[n] - A[1]`
	`2`	`+`
	`3`	`+Checking only this is suffiicent.`
	`4`	`+`
	`5`	`+------`
	`6`	`+`
	`7`	`+int main()`
	`8`	`+{`
	`9`	`+ const int NO_OF_ANTENNA = 5;`
	`10`	`+ vector <int> A(NO_OF_ANTENNA + 1);`
	`11`	`+ int max_distance;`
	`12`	`+ for(int i = 1; i <= NO_OF_ANTENNA; i++)`
	`13`	`+ {`
	`14`	`+ cin >> A[i];`
	`15`	`+ }`
	`16`	`+ cin >> max_distance;`
	`17`	`+`
	`18`	`+ int possible = (A[NO_OF_ANTENNA] - A[1] <= max_distance ? true : false);`
	`19`	`+ cout << (possible ? "Yay!" : ":(") << "\n";`
	`20`	`+`
	`21`	`+ return 0;`
	`22`	`+}`

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	`1`	`+Notice that every dish other than 1, will finish at a multiple of 10`
	`2`	`+`
	`3`	`+Round up every element to the nearest multiple of 10, apart from the one with the smallest remainder modulo i`
	`4`	`+`
	`5`	`+------`
	`6`	`+`
	`7`	`+int main()`
	`8`	`+{`
	`9`	`+ const int NO_OF_DISHES = 5;`
	`10`	`+ vector <int> A(NO_OF_DISHES + 1);`
	`11`	`+ for(int i = 1; i <= NO_OF_DISHES; i++)`
	`12`	`+ {`
	`13`	`+ cin >> A[i];`
	`14`	`+ }`
	`15`	`+`
	`16`	`+ int total_time = 0;`
	`17`	`+ int time_saved = 0;`
	`18`	`+ for(int i = 1; i <= NO_OF_DISHES; i++)`
	`19`	`+ {`
	`20`	`+ int remaining = (A[i]%10 != 0 ? 10 - A[i]%10 : 0);`
	`21`	`+ time_saved = max(time_saved, remaining);`
	`22`	`+`
	`23`	`+ total_time += A[i] + remaining;`
	`24`	`+ }`
	`25`	`+ total_time -= time_saved;`
	`26`	`+`
	`27`	`+ cout << total_time << "\n";`
	`28`	`+`
	`29`	`+ return 0;`
	`30`	`+}`

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