Zelensky congratulates Sandu on winning Moldova’s presidential election
President Volodymyr Zelensky has congratulated Maia Sandu on her victory in the presidential election in Moldova.
The head of the state wrote this on social media platform X, Ukrinform reports.
“Congratulations to Maia Sandu on winning the presidential election in Moldova. Ukraine supports the European choice of the Moldovan people and stands ready to work together to strengthen our partnership,” Zelensky wrote.
According to him, Moldovans have made a clear choice—they chose a path toward economic growth and social stability.
“Only true security and a peaceful, united Europe can guarantee each person and every family the confidence to face tomorrow with hope and certainty,” Zelensky added.
NewsMaker reports that as of the morning of November 4, the Central Election Commission of Moldova (CEC) has processed 99.5% of the ballots in the country's presidential election runoff.
Moldova’s incumbent President Maia Sandu, a candidate from the Party of Action and Solidarity, received 55.25% of the vote.
Alexandr Stoianoglo, a candidate from the Socialist Party of Moldova, received 44.75% of the vote.
The voter turnout was 54.30%. About 1.7 million people cast their ballots, including more than 328,000 outside the country.
As Ukrinform reported, Moldova held the second round of presidential elections on Sunday, November 3.
Moldovan President Maia Sandu announces her victory in the presidential election.
Photo: President’s Office