How to apply Newton polygon to discuss about the roots of a multi-variate p-adic power series ?

We know that if $f(x)=\sum_{i=0}^{\infty} a_ix^i$ be a power series over $p$-adic field, then the Newton polygon of $f(x)$ is the lower convex hull of the segments joining $(i, v(a_i))$, where $v$ is the $p$-adic valuation.

Now consider the p-adic power series $f(x_1,~x_2, \cdots, x_n)=\sum a_{h_i,~h_2, \cdots, h_n} x_1^{h_1}x_2^{h_2} \cdots x_n^{h_n}$ of $n$-variables.

Question:

How to find Newton polygon of it in the above sense ?

My Intuition:

Although, I don't have any concrete method but I have thought the following intuitions.

$(1)$ I not sure whether we can consider only one variable, say, $x_i$, and ignore the rest variablesin order to find the Newton polygon of $f(x_1,\cdots, x_n)$.

But I am thinking about the following method:

$(2)$

$(a)$ We know that $\text{Newton polygon}$ of product of two single variable power series is obtained by $\text{gluing}$ together the segments of the $\text{Newton polygon}$ of each individual power series under $\text{slope condition}$. For example, if $f(x)$ and $g(x)$ be two p-adic power series. Let $S(f)$, $S(g)$ be respectively denote the slope sets of the segments (or straight lines) of Newton polygon of $f(x)$ and $g(x)$. Assume $S(f) \leq S(g)$, then $\text{Newton polygon}$ of the product $f(x)g(x)$ is obtained by $\text{gluing}$ together the segments of $\text{Newton polygon}$ of $f(x)$ and $g(x)$.

$(b)$ So we can extend this above method for finite product of power series.

$(c)$ Now if we assume that $f(x_1,~x_2, \cdots, x_n)=f_1(x_1) \cdot f_2(x_2) \cdots f_n(x_n)$, then we can find $\text{Newton polygon}$ of each factors $f_i(x_i)$ and by the above method, joining all segments of $\text{Newton polygon}$ of each factor $f_i(x_i)$,we will get the Newton polygon of the multi-variable power series $f(x_1,~x_2, \cdots, x_n)$.

Am I thinking right way ?

Any help please.

How to apply Newton polygon to discuss about the roots of a multi-variate p-adic power series ?

We know that if $f(x)=\sum_{i=0}^{\infty} a_ix^i$ be a power series over $p$-adic field, then the Newton polygon of $f(x)$ is the lower convex hull of the segments joining $(i, v(a_i))$, where $v$ is the $p$-adic valuation.

Now consider the p-adic power series $f(x_1,~x_2, \cdots, x_n)=\sum a_{h_i,~h_2, \cdots, h_n} x_1^{h_1}x_2^{h_2} \cdots x_n^{h_n}$ of $n$-variables.

Question:

How to find Newton polygon of it in the above sense ?

My Intuition:

Although, I don't have any concrete method but I have thought the following intuitions.

$(1)$ I not sure whether we can consider only one variable, say, $x_i$, and ignore the rest variables in order to find the Newton polygon of $f(x_1,\cdots, x_n)$.

But I am thinking about the following method:

$(2)$

$(a)$ We know that $\text{Newton polygon}$ of product of two single variable power series is obtained by $\text{gluing}$ together the segments of the $\text{Newton polygon}$ of each individual power series under $\text{slope condition}$. For example, if $f(x)$ and $g(x)$ be two p-adic power series. Let $S(f)$, $S(g)$ be respectively denote the slope sets of the segments (or straight lines) of Newton polygon of $f(x)$ and $g(x)$. Assume $S(f) \leq S(g)$, then $\text{Newton polygon}$ of the product $f(x)g(x)$ is obtained by $\text{gluing}$ together the segments of $\text{Newton polygon}$ of $f(x)$ and $g(x)$.

$(b)$ So we can extend this above method for finite product of power series.

$(c)$ Now if we assume that $f(x_1,~x_2, \cdots, x_n)=f_1(x_1) \cdot f_2(x_2) \cdots f_n(x_n)$, then we can find $\text{Newton polygon}$ of each factors $f_i(x_i)$ and by the above method, joining all segments of $\text{Newton polygon}$ of each factor $f_i(x_i)$,we will get the Newton polygon of the multi-variable power series $f(x_1,~x_2, \cdots, x_n)$.

Am I thinking right way ?

Any help please.