# Elementar cell of Y2O3

I think this picture from wikipedia is wrong because I count:

\begin{align} \left. \begin{array}{ccccr} 1&×&1 &= &\ce{1~Y} \\ 6&×&\frac{1}{8} &= &\ce{\frac{3}{4}O} \end{array} \right\} \text{ gives }\ce{Y4O3} \end{align}

How would be the correct elementary cell of $\ce{Y2O3}$?

(I didn't found any correct elementary cell using google)

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