BeautifulSoup - взять из ссылки только нужные мне фрагменты

Как мне взять из ссылки "2701823" https://kaspi.kz/shop/p/birjusa-m132-seryi-2701823/?c=750000000 (ссылки будут меняться)?

from bs4 import BeautifulSoup as bs
import requests
import re

headers = {'accept': '*/*',
           'user-agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) snap Chromium/78.0.3904.108 Chrome/78.0.3904.108 Safari/537.36'}
base_url = 'https://kaspi.kz/shop/c/smartphones/?q=%3Acategory%3ASmartphones&page=1'


def kaspi_parser(base_url, headers):
    products = []
    urls = []
    urls.append(base_url)
    product_pages = []
    product_pages.append('https://kaspi.kz')
    session = requests.Session()
    request = session.get(base_url, headers=headers)
    if request.status_code == 200:
        soup = bs(request.content, 'html.parser')
        try:
            pagination = soup.find_all('a', attrs={'class': 'pagination__el'})
            count = int(pagination[-2].text)
            for i in range(1, count + 50):
                url = f'https://kaspi.kz/shop/c/smartphones/?q=%3Acategory%3ASmartphones&page={i}'
                if url not in urls:
                    urls.append(url)
        except:
            pass

    for url in urls:
        request = session.get(url, headers=headers)
        soup = bs(request.content, 'html.parser')
        divs = soup.find_all('div', attrs={'item-card__info'})
        for div in divs:
            try:
                href = div.find('a', 'ddl_product_link')['href']
                id = div.find('a', class_=re.compile('-id'))
                print(href)
                print(id)
            except:
                pass

    else:
        print('ERROR')
    return products
kaspi_parser(base_url, headers)