Character Matrix – Alternate Ends

Character Matrix – Alternate Ends: The program must accept a character matrix of size RxC as the input. The program must print the output based on the following conditions.
– The first character in the 1^st row of the matrix must be printed in the 1^st line.
– The last 2 characters in the 2^nd row of the matrix must be printed in the 2^nd line.
– The first 3 characters in the 3^rd row of the matrix must be printed in the 3^rd line.
– The last 4 characters in the 4^th row of the matrix must be printed in the 4^th line.
– Similarly, the program must print the characters of the matrix in the remaining lines as the output.

Boundary Condition(s):
2 <= R, C <= 100

Input Format:
The first line contains R and C separated by a space.
The next R lines, each contains C characters separated by a space.

Output Format:
The first R lines contain the characters based on the given conditions.

Example Input/Output 1:
Input:
6 6
a b c d e f
g h i j k l
m n o p q r
s t u v w x
y z a b c d
e f g h i j

Output:
a
kl
mno
uvwx
yzabc
efghij

Explanation:
The first character in the 1^st row is a.
The last 2 characters in the 2^nd row are k and l.
The first 3 characters in the 3^rd row are m, n and o.
The last 4 characters in the 4^th row are u, v, w and x.
The first 5 characters in the 5^th row are y, z, a, b and c.
The last 6 characters in the 6^th row are e, f, g, h, i and j.
Hence the output is 
a
kl
mno
uvwx
yzabc
efghij

Example Input/Output 2:
Input:
8 5
v f b z c
w o j E a
o F f b k
s o f x l
v n i n C
o u r r B
c p w E x
E r b l o

Output:
v
Ea
oFf
ofxl
vninC
ourrB
cpwEx
Erblo

m,n=map(int,input().split())
li=[ list(input().split()) for i in range(m) ]
for i in range(1,m,2):
    print(*li[i-1][:i],sep="")
    print(*li[i][-i-1:],sep="")
if m%2==1:
    print(*li[-1][:m],sep="")