Vatican, (IKA/VIS) – Following is the program of Benedict XVI’s apostolic trip to Croatia – scheduled for 4 – 5 June – for the National Day of Croatian Families.

The Pope will leave at 9:30am on Saturday, 4th of June, from Rome’s Fiumicino Airport and will land at Pleso International Airport in Zagreb at 11:00am.
After the welcoming ceremony, a courtesy visit will be made to Ivo Josipovic, the President of the Republic, at the presidential palace, following which the Pope will meet with Prime Minister Jadranka Kosor at the apostolic nunciature at 1:50pm.

At 6:15pm, in the Croatian National Theater of Zagreb, he will meet representatives of civil society, the political, academic, cultural, and business world, the diplomatic corps, and religious figures. At 7:30pm, at Josip Jelacic Square, he will preside over a prayer vigil with the Croatian youth.

On Sunday, 5th of June, at 10:00am, he will celebrate Holy Mass on the National Day of Croatian Families in the Zagreb Hippodrome.
At 2:00pm he will have lunch at the new offices of the Secretary of the Croatian Bishops’ Council with the Croatian and other invited bishops.
At 5:00pm he will presided over vespers with the bishops, priests, religious, and seminarians and will pray at the tomb of Blessed Aloysius Viktor Stepinac at Zagreb’s Cathedral dedicated to the Assumption of the Blessed Virgin Mary and to St. Stephen. When finished there he will meet with Cardinal Josip Bozanic, archbishop of Zagreb, at his residence, and at 7:15pm will travel to Pleso Airport where, at 7:45pm he will make the return flight to Rome. The flight is scheduled to land at Rome’s Ciampino Airport at 9:15pm.