By Linda Bordoni
Pope Francis will travel to Romania for an Apostolic Visit that will take him to the capital Bucharest, to the eastern university city of Iași, to Blaj in Transylvania and to the Marian Shrine of Sumuleu Ciuc on the outskirts of Miercurea Ciuc city, also in Transylvania.
Holy See Press Office “ad-interim” Director, Alessandro Gisotti, said the visit is scheduled to take place from 31 May to 2 June. It comes following an invitation by State Authorities and the Romanian Catholic Church.
The motto of the visit is “Let’s Walk Together.” The trip’s logo, in blue and gold, depicts a group of Romanian people walking beneath an image of Our Lady, which according to a statement from the Holy See Press Office, evokes the Virgin Mary’s care and protection of the Romanian people.
The statement also noted that Romania has often been called “the garden of the Mother of God,” which is a phrase used also by Pope St. John Paul II during his Apostolic Journey to the country.
In the footsteps of Pope St. John Paul II
Pope Francis’ visit to Romania follows exactly 20 years after Pope St. John Paul II was the first Pope to go to Romania in 1999.
The Press Office statement also notes that “The visit of Pope Francis takes up this Marian accent, inviting everyone to join forces under the protective mantle of Our Lady,” and adds that “Pope Francis has always urged for unity among different forces, for the rejection of egoisms and for the centrality of the common good”.
It concludes saying that “The Successor of Peter comes to Romania to foster unity and confirm the faith”.
Catholics in Romania make up over 4 percent of the population and are the second largest denomination after the Romanian Orthodox Church.
The Pope will be traveling to the region earlier in the month when he will visit Bulgaria and Macedonia from 5 to 7 May.