_vm.paginated не является ошибкой функции с vue-paginate

поэтому я попытался использовать плагин Vue-Paginate, однако я вижу следующую ошибку.

_vm.paginated is not a function

Вот мой код:

<template>
  <section :class = "open ? 'block': 'hidden'">
    <!-- Start Results -->
    <h1 class = "font-thin text-black text-center w-full">Cruise Results</h1>

    <section class = "flex flex-wrap">
      <paginate name = "cruises" :list = "cruises" :per = "3">
        <section class = "shadow p-4 mb-4 w-full" v-for = "cruise in paginated('cruises')" :key = "cruise.id">
          <span class = "text-md font-bold">{{ cruise.title }}</span>
          <span class = "text-sm font-semibold">{{ cruise.summary }}</span>

          <section class = "flex">
            <section class = "text-lg">
              <span class = "text-black">Inside</span>
              <br>
              <span class = "text-red">£{{ cruise.inside }}</span>
            </section>
            <span class = "pr-8"></span>
            <section class = "text-lg">
              <span class = "text-black">Balcony</span>
              <br>
              <span class = "text-red">£{{ cruise.balcony }}</span>
            </section>
          </section>
        </section>
      </paginate>
    </section>
    <paginate-links for = "cruises" :show-step-links = "true"></paginate-links>
  </section>
  <!-- Cruise Results -->
</template>

<script>
import firebase from '@/middleware/firebase'
import VuePaginate from 'vue-paginate'

const database = firebase.database()

export default {
  components: {
    VuePaginate
  },
  data: function () {
    return {
      cruises: [],
      paginate: ['cruises']
    }
  },
  mounted () {
    database.ref('cruises').on('child_added', snapshot => this.cruises.push(snapshot.val()))
  }
}
</script>

Я реализовал обе Firebase и считаю, что правильно следил за документацией.

Любая помощь здесь приветствуется.