Maia Sandu won Sunday’s presidential elections in the Republic of Moldova and will have a second term as president. After processing 99.77% of the ward minutes, Maia Sandu leads with 55.37% (928,558 votes), while Alexandr Stoianoglo has 44.63% (ie 748,506 votes). There is a significant difference of 180,052 votes between the two.
Pe the territory of the Republic of MoldovaStoianoglo obtained 51.19% (692,533 votes), while Maia Sandu obtained 48.81% (660,226 votes).
In diaspora however, Maia Sandu had an overwhelming vote, obtaining 82.78% (269,496 votes), compared to only 17.22% (56,068) for Stoianoglo.
Maia Sandu also won in the capital Chisinau with 57.38% (215,493 votes), while Alexandr Stoianoglo obtained 42.62% (160,062 votes).
It should be noted that Maia Sandu won the vote from Orhei districtthe fief of Ilan Şor – the convicted fugitive who was Russia’s main instrument in the vote-buying attempt. In Orhei, Maia Sandu obtained 52.24%, and Stoianoglo – 47.76%.
Maia Sandu also won in the districts of Străseni, Cantemir, Leova, Hîncești, Nisporeni, Ungheni.
Stoianoglo also won in Gagauzia (97.04%), Balti municipality (69.98%), Transnistrian region (79.4%), as well as in Cahul, Taraclia, Fălești, Glodeni districts.
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