Kremlin does not recognise Maia Sandu as president of Moldova
Pravda Ukraine
Dmitry Peskov, the spokesman of Kremlin ruler Vladimir Putin, has questioned the legitimacy of Moldova’s presidential election, claiming Maia Sandu could not be deemed the country’s president.
Source: Russian Kommersant, as reported by European Pravda
Details: Peskov said that the elections in Moldova were neither just nor fair, but were full of manipulation.
Quote: “As for Ms Sandu, you know that she is not, as far as we understand, the president of her country, because in the country itself, too, the majority of the population did not vote for her. And we see a very, very divided society.”
According to him, the Russian side can “objectively judge these elections” on the example of Moldovan citizens who voted in the Russian Federation.
Peskov said that due to the fact that Moldovans in the Russian Federation “were not given the opportunity to use their vote,” the election results “cannot be considered transparent”.
It is worth noting that the Moldovan authorities opened only two voting stations in Russia – both in Moscow. At each of them, 4,999 voters out of the possible 5,000 cast their vote.
Peskov also said that the residents of Transnistria “are not inclined to in any way support” the current leadership of Moldova.
Background:
- The pro-Russian Party of Socialists of Moldova, which nominated Alexandr Stoianoglo in the election, stated that it would not accept the results of voting in foreign polling sites. The socialists referred to Stoianoglo as the “real winner”.
- Former Moldovan President Igor Dodon, the socialists’ leader, referred to Sandu as “the president of the diaspora”.
- According to the vote count, Sandu was re-elected for a second term, overcoming the socialist candidate Stoianoglo by more than 10%; at foreign polling sites, more than 80% of voters voted for Sandu, although Stoianoglo received the most votes within Moldova.
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