Is there a finite $p$-group $G$ such that :

(a) $G= \langle A,x,y \rangle$, with $G/Z(G)$ has exponent $p$, and $A$ is a maximal abelian normal subgroup of $G$;

(b) the class of $G$ is equal to $p$;

(c) the subgroup $\langle B,y \rangle$ has class exactly $p$, where $B$ denotes the center of $\langle A,x \rangle$.

Thanks in advance

Is there a finite $p$-group $G$ such that :

(a) $G= \langle A,x,y \rangle$, with $G/Z(G)$ has exponent $p$, $A$ is a maximal abelian normal subgroup of $G$, and $G/A$ has order $p^2$ (thus it is elementary abelian of rank $2$);

(b) the class of $G$ is equal to $p$;

(c) the subgroup $\langle B,y \rangle$ has class exactly $p$, where $B$ denotes the center of $\langle A,x \rangle$.

Thanks in advance

Is there a finite $p$-group $G$ such that :

(a) $G= \langle A,x,y \rangle$, where $A$ is a maximal abelian normal subgroup of $G$ and $G/A$ is elemenatry abelian;

(b) the class of $G$ is equal to $p$;

(c) the subgroup $\langle B,y \rangle$ has class exactly $p$, where $B$ denotes the center of $\langle A,x \rangle$.

Thanks in advance

Is there a finite $p$-group $G$ such that :

(a) $G= \langle A,x,y \rangle$, with $G/Z(G)$ has exponent $p$, and $A$ is a maximal abelian normal subgroup of $G$;

(b) the class of $G$ is equal to $p$;

(c) the subgroup $\langle B,y \rangle$ has class exactly $p$, where $B$ denotes the center of $\langle A,x \rangle$.

Thanks in advance