VATICAN CITY, JUL 11, 1996 (VIS) - Made public today was the itinerary for Pope John Paul's 73rd foreign pastoral trip on September 6 and 7 to Hungary. During his visit to Budapest, Pannonhalma and Gyor, he will deliver one homily and five speeches.
The Pope is scheduled to leave Rome at 8:30 a.m. on Friday, September 6, arriving in Budapest just over 2 hours later, where he will speak at the airport. He will transfer to Pannonhalma by helicopter and have a private audience with the president of the republic in the archabbey. At 7 p.m. there will be a celebration of vespers on the occasion of the millennium of the foundation of the Archabbey of Pannonhalma.
On September 7, the Holy Father will deliver a homily during a Eucharistic concelebration in Ipari Park in Gyor. At 1 p.m. he will meet with and address members of the Hungarian Bishops' Conference in the bishopric of Gyor where, three hours later he is scheduled to meet with the prime minister. He will then go to the cathedral and address representatives of the diocese.
Following a helicopter transfer to Budapest and a farewell speech at the Ferihegy I Airport, he will leave for Rome. Arrival is scheduled for 9:30 p.m.