Seating Arrangement NIT AP

Students of NIT-AP are very obedient . They come to class daily and sit according to seats pre assigned to them. Every Student Ax is pre assigned a seat Sx ,which is unique, for the entire class. There is no chance of alteration .Benches are arranged in classroom in (rows X column ) fashion.
As we all know ,there is a shortage of benches in NIT-AP. So three students are requested to sit on one bench corresponding to their seat id Sx .
But Seat ids are assigned with some logic. For every odd or even numbered rows , benches are arranged in ascending order from left to right. 
On each Bench , on an odd numbered row ,seat ids are arranged in ascending order from left to right .
But in case of even numbered rows On each Bench, seat ids are arranged in descending order from left to right .
However in each case (whether odd or even ) benches are arranged in ascending order of their bench ids from left to right.




Some of the students may be bright and absent on some days. For them the seat id Sx against their name remains vacant for the entire session.
A pair (as showed in diagram above ) is said to be lucky if he has a immediate neighbor either on the left or right on the immediate neighbor bench. For e.g.
S13 and S18 make 1 lucky pair (provided both are present )
||y S6 and S7 make 1 lucky pair (provided both are present ) .
Your task is to print
1) the total no of lucky pairs 2) Class sitting pattern.
______________________________________________________________________________
Input
n-> Total no of Sids against students;
C-> columns, maximum columns of benches to be in each row;
ab-> No of students who are absent;
abs[ab]->an array having Sids of absent students.
______________________________________________________________________________
Output
Task 1-> Seat Arrangement.
Task 2-> No. of lucky pairs.