Bucharest, Oct 15 /Agerpres/ - Foreign Minister Teodor Melescanu will attend on Monday the meeting of the Foreign Affairs Council (CAE), which will take place in Luxembourg.
According to a Foreign Ministry (MAE) press release issued for AGERPRERS on Sunday, the agenda of the meeting includes topical issues such as Iran, the Democratic People's Republic of Korea and human rights.
During a working luncheon, ministers will exchange views on Turkey and the regional situation.
On the CAE sidelines, Lithuanian Foreign Minister Linas Linkevicius will host a working breakfast, to which was invited Vladimir Kara-Murza, deputy chairman of the Open Russia movement, the MAE also informs. AGERPRES (RO - editor: Florin Marin; EN - editor: Bogdan Gabaroi)