Proof using natural deduction of $\small\forall x~[A(x)\to\exists y~[B(y)\land C(y,x)]] \vdash\forall x~[\exists y~[A(x)\to(B(y)\land C(y,x))]]$

Question

I am trying to construct a formal proof of the argument: $$\forall x~\Big[A(x)\to\exists y~\big[B(y)\land C(y,x)\big]\Big] ~\vdash~\forall x~\Big[\exists y~\big[A(x)\to(B(y)\land C(y,x))\big]\Big]$$ using natural deduction, but I'm kind of stuck.