Get Wikipedia page ID from WikiData ID

I want to get Wikipedia page ID from Wikidata ID from Wikidata Query Service in a single SPARQL query.

First SPARQL query to get Wikidata ID (wd:Q3131) from title page: "Alpes-de-Haute-Provence"

SELECT ?item ?itemLabel ?itemDescription WHERE {
            ?url rdf:type schema:Article;
            schema:about ?item; schema:name ?name ;
            schema:isPartOf <https://fr.wikipedia.org/>;
            schema:name "Alpes-de-Haute-Provence"@fr.
          SERVICE wikibase:label {
            bd:serviceParam
            wikibase:language "fr,[AUTO_LANGUAGE],en"
          }
}

Second SPARQL query to get Wikipedia page ID from Wikidata ID

SELECT ?pageid WHERE {
    VALUES (?item) {(wd:Q3131)} 
    [ schema:about ?item ; schema:name ?name ;
      schema:isPartOf <https://fr.wikipedia.org/> ]
     SERVICE wikibase:mwapi {
         bd:serviceParam wikibase:endpoint "fr.wikipedia.org" .
         bd:serviceParam wikibase:api "Generator" .
         bd:serviceParam mwapi:generator "allpages" .
         bd:serviceParam mwapi:gapfrom ?name .
         bd:serviceParam mwapi:gapto ?name .
         ?pageid wikibase:apiOutput "@pageid" .
    }
}

Is possible to do it in a single SPARQL query?

Solution proposed by @UninformedUser

SELECT ?item ?itemLabel ?itemDescription ?pageid WHERE {             
  ?url rdf:type schema:Article;             
  schema:about ?item; 
  schema:name ?name ;             
  schema:isPartOf <https://fr.wikipedia.org/>;             
  schema:name "Alpes-de-Haute-Provence"@fr.           
  SERVICE wikibase:label {             
    bd:serviceParam             
    wikibase:language "fr,[AUTO_LANGUAGE],en"           
  }
  SERVICE wikibase:mwapi {
    bd:serviceParam wikibase:endpoint "fr.wikipedia.org" .          
    bd:serviceParam wikibase:api "Generator" .          
    bd:serviceParam mwapi:generator "allpages" .          
    bd:serviceParam mwapi:gapfrom ?name .          
    bd:serviceParam mwapi:gapto ?name .          
    ?pageid wikibase:apiOutput "@pageid" .     
  } 
}