MOSCOW, October 9. /TASS/. Russian President Vladimir Putin will meet with his Belarusian counterpart Alexander Lukashenko and give him Russia’s highest award, the Order of St. Andrew the Apostle the First-Called, Kremlin aide Yuri Ushakov has told reporters.
The Kremlin official said that the bilateral Russian-Belarusian meeting will take place "in the context of the CIS summit," held in Moscow on October 8. It will also come in continuation of the informal CIS summit that took place in the Kremlin the day before.
Putin and Lukashenko enjoy a strong friendship, the Kremlin has said. The two leaders meet often and they hold telephone conversations on a regular basis. Both of them excel in protecting the interests of their countries, which enables them to protect the interests of the Union State as well, Russian Presidential Spokesman Dmitry Peskov has noted. Wednesday’s meeting will be their eighth overall in 2024.
Congratulating Lukashenko in his birthday on August 30, the Russian leader said that his good, friendly relations with Lukashenko had stood the test of time. "And of course, I am confident that we will continue doing our best to strengthen the fruitful cooperation between Russia and Belarus," the Russian leader added.
The Belarusian leader was awarded "for outstanding achievements in developing the allied relationship and strategic partnership" between Russia and Belarus as well as "for boosting friendly ties between the Russian and Belarusian peoples," the Kremlin said on its website. Lukashenko’s major personal contribution to the establishing and effective functioning of the Union State was also emphasized.