Aleksandrowo (Sejny Kūn)
Aleksandrowo sī chi̍t ê tī Pho-lân Kiōng-hô-kok Podlaskie Séng Sejny Kūn Krasnopol Kong-siā ê chng-thâu.[1]
| Aleksandrowo | |
| — Chng-thâu — | |
  Aleksandrowo |
|
| Keng-hūi-tō͘: 54°5′47″N 23°9′29″E / 54.09639°N 23.15806°E | |
| Kok-ka |  Pho-lân
|
|---|---|
| Séng | Podlaskie |
| Kūn | Sejny |
| Kong-siā | Krasnopol |
| Sî-khu | CET (UTC+1) |
| - Jo̍at--lâng | CEST (UTC+2) |
Chham-oa̍t siu-kái
Chham-khó siu-kái
- ↑ "Central Statistical Office (GUS) - TERYT (National Register of Territorial Land Apportionment Journal)" (ēng Pho-lân-gí). 2008-06-01.