How to make transition from def back(call): to search(message)?

Question

Python

There is a telegram bot with Inline buttons, I need to be redirected to def search(message) after pressing the button:

@bot.message_handler(content_types=['text'])
def echo_all(message):
    pass
def search(message):
    print("молодец")
#СЮДА НУЖНО ПЕРЕЙТИ ИЗ call.data == '16-20'
@bot.callback_query_handler(func=lambda call: True)
def callback_query(call):
elif call.data == "search":
        bot.delete_message(call.message.chat.id, call.message.message_id)
        kb = InlineKeyboardMarkup(row_width = 1)
        kb.add(InlineKeyboardButton("16-20", callback_data="16-20"))
        kb.add(InlineKeyboardButton("21-25", callback_data="21-25"))
        kb.add(InlineKeyboardButton("26-30", callback_data="26-30"))
        kb.add(InlineKeyboardButton("30 +", callback_data="30 +"))
        bot.send_message(id, "Выберите возрастную категорию:",reply_markup = kb)

elif call.data == '16-20':
      #ТУТ НУЖНО ПЕРЕЙТИ НА def search(message)

Tried search(message), search(call.message) doesn't work, please help.
bot.register_next_step_handler(call.message, search) Not suitable, I need the function to be called immediately