Aleksandrów (Garwolin Kūn)

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Aleksandrów
—  Chng-thâu  —
Aleksandrów is located in Pho-lân
Aleksandrów
Aleksandrów
Keng-hūi-tō͘: 51°48′56″N 21°32′42″E / 51.81556°N 21.54500°E / 51.81556; 21.54500
Kok-ka  Pho-lân
Séng Masovia
Kūn Garwolin
Kong-siā Łaskarzew
Sî-khu CET (UTC+1)
 - Jo̍at--lâng CEST (UTC+2)

Aleksandrów sī chi̍t ê tī Pho-lân Kiōng-hô-kok Masovia Séng Garwolin Kūn Łaskarzew Kong-siā ê chng-thâu.[1]

Chham-oa̍t[siu-kái | kái goân-sí-bé]

Chham-khó[siu-kái | kái goân-sí-bé]

  1. ↑ "Central Statistical Office (GUS) - TERYT (National Register of Territorial Land Apportionment Journal)" (ēng Pho-lân-gí). 2008-06-01. 
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