Question: Party budget planning - complexPrompt the user to enter their current budget and the number of people who need to share a meal.You are ordering food from a restaurant that has two menu items:Tacos $4Empanadas $3Calculate and print out all the options of meals where the budget is used entirely(where possible) and each person has an equal number of items to eat. They may eat different things, but everyone gets the same number of food items, e.g. 2 tacos or 1 empanada and 1 taco or 2 empanadas.Hints: Here are some things that might help you in your solution.If total items purchased does not equally divide by the number of diners, then abandon that combination.It may help to calculate the max number of empanadas that the budget can buy and also the max number of tacos that the budget could buy. These can be thought of as upper bounds in your loops.In your solution, keep track of items per personAlso, keep track of total number of viable solutions, as that might help you determine when there is no solution.My solution is printing "You cannot buy a meal with these options." multiple times even when it is possible to buy a meal. Also my solution does show "Meal options:" before printing all of the alternatives. I've attached the outcome that the program should be displaying and my current code. Enter budget for meal: $100How many people are eating: 6Meal options:$98 buys 22 empanadas and 8 tacos with $2 change.5 items per person.$99 buys 21 empanadas and 9 tacos with $1 change.5 items per person.$100 buys 20 empanadas and 10 tacos without change. 5 items per person.orEnter budget for meal: $33How many people are eating: 15You cannot buy a meal with these options. def get_possible_meals ( budget, num_persons):tacos_cost = 4empanandas_cost= 3budget // tacos_costbudget // empanandas_costmax tacos =max_empanadasfeasible = Falsefor num_tacos in range (max_empanadas + 1):num_empanadasif num_empanadas 0 and not (num_empanadas + num_tacos) % num_persons:feasible = True(budgetnum_tacos * tacos_cost) // empanandas_costtotal_costnum_tacos * tacos_cost + num_empanadas * empanandas_cost%3Dchangebudgettotal_cost%3Dprint(f'${total_cost} buys {num_empanadas} empanadas and {num_tacos} tacos with ${change} change.')print(f'{(num_empanadas + num_tacos) // num_persons} items per person.')if not feasible:print(f'You cannot buy a meal with these options.')def main() :budgetint(input('Enter budget for meal: $'))num_personsint(input('How many people are eating: '))get_possible_meals(budget, num_persons)main__':ifname==main()→ Enter budget for meal: $100How many people are eating: 6You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.$98 buys 22 empanadas and 8 tacos with $2 change.5 items per person.

Question

Question: Party budget planning - complexPrompt the user to enter their current budget and the number of people who need to share a meal.You are ordering food from a restaurant that has two menu items:Tacos $4Empanadas $3Calculate and print out all the options of meals where the budget is used entirely(where possible) and each person has an equal number of items to eat. They may eat different things, but everyone gets the same number of food items, e.g. 2 tacos or 1 empanada and 1 taco or 2 empanadas.Hints: Here are some things that might help you in your solution.If total items purchased does not equally divide by the number of diners, then abandon that combination.It may help to calculate the max number of empanadas that the budget can buy and also the max number of tacos that the budget could buy. These can be thought of as upper bounds in your loops.In your solution, keep track of items per personAlso, keep track of total number of viable solutions, as that might help you determine when there is no solution.My solution is printing &#34;You cannot buy a meal with these options.&#34; multiple times even when it is possible to buy a meal. Also my solution does show &#34;Meal options:&#34; before printing all of the alternatives. I've attached the outcome that the program should be displaying and my current code. Enter budget for meal: $100How many people are eating: 6Meal options:$98 buys 22 empanadas and 8 tacos with $2 change.5 items per person.$99 buys 21 empanadas and 9 tacos with $1 change.5 items per person.$100 buys 20 empanadas and 10 tacos without change. 5 items per person.orEnter budget for meal: $33How many people are eating: 15You cannot buy a meal with these options. def get_possible_meals ( budget, num_persons):tacos_cost = 4empanandas_cost= 3budget // tacos_costbudget // empanandas_costmax tacos =max_empanadasfeasible = Falsefor num_tacos in range (max_empanadas + 1):num_empanadasif num_empanadas > 0 and not (num_empanadas + num_tacos) % num_persons:feasible = True(budgetnum_tacos * tacos_cost) // empanandas_costtotal_costnum_tacos * tacos_cost + num_empanadas * empanandas_cost%3Dchangebudgettotal_cost%3Dprint(f'${total_cost} buys {num_empanadas} empanadas and {num_tacos} tacos with ${change} change.')print(f'{(num_empanadas + num_tacos) // num_persons} items per person.')if not feasible:print(f'You cannot buy a meal with these options.')def main() :budgetint(input('Enter budget for meal: $'))num_personsint(input('How many people are eating: '))get_possible_meals(budget, num_persons)main__':ifname==main()→ Enter budget for meal: $100How many people are eating: 6You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.$98 buys 22 empanadas and 8 tacos with $2 change.5 items per person.

Accepted Answer

You are ordering food from a restaurant that has two menu items:

Tacos $4

Empanadas $3