( Javno ) - In a regular report on the journeys and meetings of American President George W. Bush due to this week, the White House announced that the president will arrive in Zagreb on Friday April 4, at 18:25, when, according to their schedule, a welcoming ceremony will be held at the Croatia President's Office.
Although the schedule of President Bush's visit is strictly kept secret by the Croatian authorities, due to, they say, security reasons, the White House obviously has no need to keep such data secret. Considering the way Bush has to go to get to the President's Office, he will probably land at the airport between half and hour and one hour earlier, so we can expect the traffic to be blocked between 5 and 6:30 p.m.
According to the published schedule, the American president with meet the Croatian President Stjepan Mesic at the President's Office, and the meeting with last an hour. At 19:50, Mr and Mrs Bush will take part in a dinner for guests, to be held in the President's Office.
On Saturday morning, Bush will meet the Croatian Prime Minister, Ivo Sanader at the cabinet's palace at 18:40, and he will hold his speech addressed to the citizens an hour later in Markov Square. At 11:50, he will take pictures with the representatives of the countries who received the NATO letter of call in Vila Prekrizje, and will have a business lunch with them.
It is still not known where Bush will go after the lunch, and his next meeting will take place at 20:00 in Russia with Valdimir Putin.