Why $\mathbb E[X_i^2\mathbb E[B_{t_i}-B_{t_i-1}\mid\mathcal F_{t_i-1}]]=\mathbb E[X_i^2]\mathbb E[B_{t_i}-B_{t_i-1}\mid \mathcal F_{t_i-1}]$?

Question

Let $(B_t)_t$ a brownian motion adapted to the filtration $(\mathcal F_t)_t$, and $X_i$ a r.v. $\mathcal F_{t_i-1}$-measurable s.t. $\mathbb E[X_i^2]<\infty$. Why $$\mathbb E\Big[\mathbb E[X_i^2(B_{t_i}-B_{t_i-1})\mid \mathcal F_{t_i-1}]\Big]=\mathbb E\Big[X_i^2\mathbb E[B_{t_i}-B_{t_i-1}\mid\mathcal F_{t_i-1}]\Big]=\mathbb E[X_i^2]\mathbb E[B_{t_i}-B_{t_i-1}\mid \mathcal F_{t_i-1}]\ \ ?$$ The first equality comme from the fact that $X_i^2$ is $\mathcal F_{t_i-1}$ measurable. But I don't understand the second one.