Django передача данных из шаблона

Пытаюсь передать переменную из шаблона в url

<li class="menu__item menu__item_header">
                            <a href="{% url 'decision:room' rmslg='kitchen'%}" class="menu__link">Жилые интерьеры</a>
                    </li>

Но он не воспринимает ее как переменную

Page not found (404)
Request Method: GET
Request URL:    http://127.0.0.1:8000/decision/livingrooms/%3Cslug:room.slug%3E/
Using the URLconf defined in caparol_center_spb_decision.urls, Django tried these URL patterns, in this order:

admin/
[name='index']
services/ [name='services']
decision/ livingrooms/ [name='livingrooms']
decision/ livingrooms/<slug:rmslg>/ [name='room']
decision/ livingrooms/<slug:rmslg>/<slug:stslg>/ [name='style']
news/
^media/(?P<path>.*)$
The current path, decision/livingrooms/<slug:room.slug>/, didn't match any of these.

urls.py

from django.urls import path
from .views import ContactView
from . import views

app_name = 'decision'

urlpatterns = [
    path('livingrooms/', ContactView.as_view(), name='livingrooms'),
    path('livingrooms/<slug:rmslg>/', views.room, name='room'),
    path('livingrooms/<slug:rmslg>/<slug:stslg>/', views.style, name='style'),
]

views.py

from django.shortcuts import render
from .forms import EmailForm
from django.views.generic.edit import FormView
from .models import Room,Style


class ContactView(FormView):
    template_name = 'decision/residentialInteriors.html'
    form_class = EmailForm
    success_url = 'decision/residentialInteriors.html'



    def form_valid(self, form):
        print("success")

def room(request, rmslg):
    styles = Room.objects.get(slug=rmslg).styles.all()
    return render(request, 'decision/residentialInteriors.html', {"styles": styles})


def style(request, stslg):
    style = Style.objects.get(slug=stslg)
    return render(request, 'decision/residentialInteriors.html', {"style": style})