Match the starting materials given in Column-$I$ with the products formed by these (Column-$II$) in the reaction with $HI$.

Column-$I$	Column-$II$
$(A)$ $CH_3-O-CH_3$	$(1)$ $C_6H_5OH + CH_3I$
$(B)$ $(CH_3)_2CH-O-CH_3$	$(2)$ $(CH_3)_3C-I + CH_3OH$
$(C)$ $(CH_3)_3C-O-CH_3$	$(3)$ $C_6H_5I + CH_3OH$
$(D)$ $C_6H_5-O-CH_3$	$(4)$ $CH_3OH + CH_3I$
	$(5)$ $(CH_3)_2CHOH + CH_3I$
	$(6)$ $(CH_3)_2CH-I + CH_3OH$
	$(7)$ $(CH_3)_3COH + CH_3I$

(A-4, B-5, C-2, D-1) The reaction of ethers with $HI$ follows $S_N2$ or $S_N1$ mechanisms depending on the nature of the alkyl groups.
$(A)$ $CH_3-O-CH_3 + HI \rightarrow CH_3OH + CH_3I$ (Matches with $4$).
$(B)$ $(CH_3)_2CH-O-CH_3 + HI \rightarrow (CH_3)_2CH-I + CH_3OH$ (The $S_N2$ mechanism occurs at the less hindered methyl group,but here the secondary carbon is more reactive towards $S_N1$ or $S_N2$ depending on conditions; typically,$CH_3I$ and $(CH_3)_2CHOH$ are formed,matching with $5$).
$(C)$ $(CH_3)_3C-O-CH_3 + HI \rightarrow (CH_3)_3C-I + CH_3OH$ (The $S_N1$ mechanism occurs at the tertiary carbon,matching with $2$).
$(D)$ $C_6H_5-O-CH_3 + HI \rightarrow C_6H_5OH + CH_3I$ (Cleavage occurs at the $O-CH_3$ bond because the $C-O$ bond to the phenyl ring has partial double bond character,matching with $1$).
Thus,the correct matching is: $A-4, B-5, C-2, D-1$.