Inequality between two random walks if comparison of moment of step is known

Question

Let $B_i =\pm 1$ be i.i.d. Bernoulli r.v. with probability $1/2$ each. Let $X_i \geq 0$ be i.i.d. r.v.'s, and independent to $B_i$'s. Let $Y_i \geq 0$ be i.i.d. r.v.'s, and independent to $B_i$'s. Suppose, for $q=1,2,3,\dotsc$, $$\mathbb{E} (X_i^q) \leq \mathbb{E} (Y_i^q) <\infty$$ Define $$\tilde{X}_N :=\sum_{i=1}^N X_i B_i$$ $$\tilde{Y}_N :=\sum_{i=1}^N Y_i B_i$$ Prove or disprove: for $q=1,2,3,\dotsc$, $$\mathbb{E} (|\tilde{X}_N|^q) \leq \mathbb{E} (|\tilde{Y}_N|^q)?$$ Instinct: It seems very plausible to be true. Since each step is symmetric, the larger the step is, the larger the total distance is, and $x^q$ (which is convex) seems even to make the difference larger. However, I do not know how to break the absolute value of $|\tilde{X}_N|^q$. We can't use Jensen inequality "on both sides" (which does not make sense).