I am trying to prove that for type $A$ , rational smoothness of Schubert varieties implies smoothness.

So suppose we are in Type $A_{n-1}$, so let $G=Sl(n,\mathbb C)$, $B=$ the group of upper triangular matrices in $G$ , $T=$ group of diagonal matrices in $G$. $W=N_G(T)/T \cong \mathfrak S_n$. Let $R=\{e_i -e_j : 1\le i, j\le n, i\ne j\}$ be our root system and $R^+=\{e_i-e_j : 1\le i<j\le n\}$ be positive root system.

Let $\le$ be the Bruhat order on $W$.

For $w\in W$ , let $\Gamma(w)$ be the Bruhat graph of $w$, whose vertex set is $\{u \in W : u\le w\}$ and between any two vertices $x$ and $y$, there is a edge if and only if $x=ys$ for some reflection $s$ (not necessarily simple).

Now suppose it is given that every vertex of the Bruhat graph $\Gamma (w)$ has degree equal to $l(w)$ (i.e. equivalently the Schubert variety $X(w)$ is rationally smooth ). Then how to show that

$l(w)=$ # $ \{\alpha \in R^+| s_{\alpha} \le w\}$ ?

[Note that if I can show my claim then I am done because $l(w)=\dim X(w)$ and dimension of the tangent space of $X(w)$ at $e_{id}$ is $\dim T(w,e_{id})=$# $ \{\alpha \in R^+| s_{\alpha} \le w\}$, so the claim would imply $X(w)$ is smooth at $e_{id}$ , hence smooth.]