PROGRAM OF POPE'S TRIP TO BOSNIA-HERZEGOVINA

VATICAN CITY, JUN 13, 2003 (VIS) - Made public today was the program of the Pope's trip to Bosnia-Herzegovina on June 22, where he will beatify Servant of God Ivan Merz.

The Holy Father will leave Rome at 8:15 a.m., arriving at Banja Luka's international airport at 9:40 a.m. After a welcome ceremony, he is scheduled to meet with the collegial presidency of the country. At 11:30 a.m. he will celebrate Mass and beatify Servant of God Ivan Merz on a plain in front of the Convent of the Most Holy Trinity of the Order of the Friars Minor on a hillside in Petricevac. He will also say a few words before praying the Angelus.

At 2:30 p.m. the Pope will eat lunch with bishops from Bosnia-Herzegovina and with cardinals and bishops of the entourage in the bishop's residence in Banja Luka. At 5:30 p.m. he is scheduled to receive the president of the Serbian Republic and the president of the Federation of Bosnia-Herzegovina and a half an hour later he will meet with the country's inter-religious council. At 6:30 p.m. he will make a private visit to the Catholic cathedral in Banja Luka. At 7:15 p.m., after a farewell ceremony in the airport, the Pope will return to Rome where he is scheduled to arrive at 9:00 p.m.

JPII-TRIP;PROGRAM BOSNIA-HERZEGOVINA;...;...;VIS;20030613;Word: 230;