what is the product of this reaction and mechanism?

![picture from national chemistry Olympiad]


[![enter image description here](https://i.stack.imgur.com/yEOtk.png)](https://i.stack.imgur.com/yEOtk.png)


I cant get the mechanism.
Answer shows there will be formation of keto group.
But how?


As I can understand the $\ce{H^+}$ is going to attack the $\ce{O}$ and can remove the $\ce{CH3}$ groups forming two adjacent alcohols which may join to form keton.
I'm not sure.




The mechanism to convert acetal to ketone in acid based hydrolysis is in following steps-


* Protonation of (1) to make OCH3 a good leaving group forming (2)
* O lone pair allows LG to leave forming oxonium ion and alcohol(3)
* Water nucleophile attacks C of C=O (3) to give (4)
* Base removes proton to give (5).
* Reprotonates (5) to make OCH3 a better leaving group(6).
* CH3OH leaves to give (7)
* Base deproronates to give ketone (8)


**Note :In this scheme, the base, B:, could be R-O-R, R-OH, H2O or the conjugate base of the acid catalyst**


[![enter image description here](https://i.stack.imgur.com/cLwtY.png)](https://i.stack.imgur.com/cLwtY.png)


**References:**


* <http://www.chem.ucalgary.ca/courses/353/exams/3513/353w05/353fin05me.html>
* <https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/aldket1.htm>


 2

Yes, the oxigen is protonated, water enters, methanol leaves and another $\ce{H^+}$ is released.
Then the oxigen of the other $\ce{CH3O}$ group is protonated, the keto group is formed while the methanol leaves and $\ce{H^+}$ is released.


Edit:


The first part results in:


[![enter image description here](https://i.stack.imgur.com/IdFnH.png)](https://i.stack.imgur.com/IdFnH.png)


and $\ce{CH3OH}$.


[![enter image description here](https://i.stack.imgur.com/t6h6n.gif)](https://i.stack.imgur.com/t6h6n.gif)


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