Prove $card(\{x\in V_N:h_N(x)\ge2\lambda\sqrt{g}\log(N)\})=N^{2(1-\lambda^2)+o(1)}$ with $|V_N|=N^2$ and $h_N(x)$ being i.i.d $\mathcal{N}(0,g\log N)$

Question

For $ N\ge 1$, set $V_N=[1,N]^2\cap\mathbb{Z}^2$. Suppose that $\{h_N(x):x\in V_N\}$ are i.i.d Gaussian variables with mean $0$ and variance $g\log(N)$ where $g>0$ is a constant. For $\lambda\in(0,1)$, denote by $L_N(\lambda)$ the cardinality of the set $\{x\in V_N:h_N(x)\ge2\lambda\sqrt{g}\log(N)\}$. Show that $$L_N(\lambda)=N^{2(1-\lambda^2)+o(1)} $$ where $o(1)\rightarrow0$ in probability as $N\rightarrow\infty$. In other words, show that for any $\epsilon\ge0$, we have $$\mathbb{P}[L_N(\lambda)\ge N^{2(1-\lambda^2)+\epsilon}]\rightarrow0, \ \mathbb{P}[L_N(\lambda)\le N^{2(1-\lambda^2)-\epsilon} ]\rightarrow0, \ as~N\rightarrow\infty.$$ $$$$ For this problem we can firstly check the easy case (e.g. $\lambda=1$), in which we can use the the same method in Prove that the maximum of $n$ independent standard normal random variables, is asymptotically equivalent to $\sqrt{2\log n}$ almost surely.. However the same method seems not working for this stronger case, and it seems not easy to derive the order of this cardinality. So are there any suggestions to derive the order bound of $L_N(\lambda)$? Thanks!