Zelensky to participate in food security summit in New York

On Tuesday, the President of Ukraine, Volodymyr Zelensky, will participate in the food security summit, which will be held as part of the high-level week of the 77th session of the UN General Assembly.

The event was organized on the initiative of the United States and the African Union, Ukrinform reports.

It will be held from 14:00 to 16:00 EST (3 21:00 to 23:00 Kyiv time). Zelensky's speech is expected in video conference mode.

The Prime Minister of Ukraine, Denys Shmyhal, will participate in the summit's work. He will also hold several bilateral meetings with the heads of governments of partner countries and international financial organizations.

Reference

On Tuesday, the general debate of the 77th session of the UN General Assembly opens, where leaders of states, heads of government, and heads of foreign policy departments of UN member states will speak. Ukraine aims to expand cooperation with foreign countries at the expense of Africa, Asia, and Latin America, as well as to develop mechanisms to hold russia accountable for the crime of aggression.

President Volodymyr Zelensky will speak at the general debate in video mode on Wednesday, September 21, during the evening session of the UN General Assembly.