rt.representation theory - Do Jones-Wenzl idempotents lift to anything interesting in the Hecke algebra?

Background

Inside the Temperley-Lieb algebra $TL_n$ (with loop value $delta=-[2]$ and standard generators $e_1,ldots,e_{n-1}$), the Jones-Wenzl idempotent is the unique non-zero element $f^{(n)}$ satisfying
$$ f^{(n)}f^{(n)} = f^{(n)} quad textrm{and} quad e_i;f^{(n)} = 0 = f^{(n)}e_i quad textrm{for each } i.$$
Consider the Iwahori-Hecke algebra $mathcal{H}_n$, $nge3$, normalized so that $(T_i-q)(T_i+q^{-1})=0$, where $q$ is generic. Let $mathcal{I}$ be the two-sided cellular ideal generated by canonical basis element
$$C_{121} = T_1T_2T_1-qT_1T_2-qT_2T_1+q^2T_1+q^2T_2-q^3.$$
The assignment $mathcal{H}_n rightarrow TL_n$ given by $T_i mapsto e_i + q$ is a surjective $mathbb{C}(q)$-algebra homomorphism with kernel $mathcal{I}$.

We can lift the generators $e_i$ in $TL_n$ to the Kazhdan-Lusztig elements $C_i=T_i-q in mathcal{H}_n$. In fact, we have $C_{121} = C_1C_2C_1 - C_1$, hence the relation down below. Rescaling a bit, $E=-frac{1}{[3]!}C_{121}$ is an idempotent, corresponding to the partition $(1,1,1)$. Actually, all of the primitive idempotents in $mathcal{H}_n$ that correspond to Young diagrams with more than two rows live in the ideal $mathcal{I}$.

Now, any preimage of $f^{(n)}$ in the Hecke algebra (call it $F^{(n)}$) satisfies
$$F^{(n)}F^{(n)} equiv F^{(n)} quad textrm{and} quad C_iF^{(n)} equiv 0 equiv F^{(n)}C_i quad (operatorname{mod} mathcal{I})$$

Question

Can we choose $F^{(n)}$ to be an idempotent in $mathcal{H}_n$?

When $n=2$, the map is an isomorphism and we have no choice.
$$F^{(2)} = frac{1}{[2]}(T_1+q^{-1}),$$
which projects onto the $q$-eigenspace for $T_1$. In other words, it is the idempotent corresponding to the partition $(2)$.