Ice to cool a drink

We want to cool $0.30\mathrm{kg}$ of a drink, which can be treated as water, from $25{}^{\circ }\mathrm{C}$ to $5.0{}^{\circ }\mathrm{C}$ by adding ice at $0{}^{\circ }\mathrm{C}$. Neglecting exchanges with the outside, compute the mass of ice needed ($c_a=4186J/(\mathrm{kg}\cdot \mathrm{K})$, ${\lambda }_f=3.33\cdot 10^{5}J/kg$).