If, for all $k\in\mathbb{N}$, $(x_{i}^{k})_{i=1}^{\infty}\in X^{\mathbb{N}}$ is normalized and $M$-basic and if, in addition, for all $k\leq i_{1}<i_{2}<\ldots$ the diagonal sequence $(x_{i_{k} }^{k})_{k=1}^{\infty}\in X^{\mathbb{N}}$ is $M$-basic, then $(x_{i}^{k})_{i=1,k\in\mathbb{N}}^{\infty}$ is said to be an $M$-basic array. Here is a definition of an asymptotic model of a Banach space $X$.

Let $(x^{k}_{i})_{i=1,k\in\mathbb{N}}^{\infty}$ be an $M$-basic array in $X$ and let $(v_{i})_{i=1}^{\infty}$ be a $1$-spreading normalized basis for a Banach space $(V,\|\cdot\|_{V})$. If there exist positive real numbers $\varepsilon_{N}\downarrow 0$ such that for all \begin{equation} \left\vert \left\Vert\sum_{k=1}^{N}\lambda_{k}x^{k}_{i_{k}}\right\Vert-\left\Vert\sum_{i=1}^{N}\lambda_{i}v_{i}\right\Vert_{V}\right\vert<\varepsilon_{N} \end{equation} for all $N\leq i_{1}<\ldots <i_{N}$ and for all scalars $(\lambda_{i})_{i=1}^{N}\in[-1,1]^{N}$, then $(v_{i})_{i=1}^{\infty}$ is said to be an asymptotic model of $X$ generated by $(x^{k}_{i})_{i=1,k\in\mathbb{N}}^{\infty}$.

My question is the following: does any sub-array of the form $(x_{i}^{k_{j}})_{i=1,j\in\mathbb{N}}^{\infty}$ where $k_{1}<k_{2}<\ldots$ generate the same asymptotic model $(v_{i})_{i=1}^{\infty}$ with respect to the same decreasing sequence $\varepsilon_{N}\downarrow 0$?