Getting 3 string before and after of each characters using awk - bash

Question

What I have expected is the output like below:
[before character h is null and assign with '#". After character h
are "e","l","l".]
[before character e is "h". After character e are
"l","l","o".]
[before character l are "h" and "e". After character l
are "l" and "o".]
[before character l are "h" and "e". After character
l are "l" and "o".]
[before character l are "h","e","l". After
character l is "o".]
[before character o are "e","l","l". After
character o is null and assign with '#".]
# # # h e l l
# # h e l l o
# h e l l o #
h e l l o # #
e l l o # # #
# # # w o n d
# # w o n d e
# w o n d e r
w o n d e r f
o n d e r f u
n d e r f u l
d e r f u l #
e r f u l # #
r f u l # # #
Input file:
h e l l o
w o n d e r f u l
Code:
awk -v s1="# # #"
'BEGIN{v=length(s1)}
{$0=s1 $0 s1;num=split($0, A,"");
for(i=v+1;i<=num-v;i++){
q=i-v;p=i+v;
while(q<=p){
Q=Q?Q OFS A[q]:A[q];q++
};
print Q;Q=""
}
}' InputFile
But the result I got is:
# # # h e l
# # h e l l
# # h e l l
# h e l l o
# h e l l o #
h e l l o #
e l l o # #
e l l o # #
l l o # # #
# # # w o n
# # w o n d
# # w o n d
# w o n d e
# w o n d e
w o n d e r
o n d e r
o n d e r f
n d e r f
n d e r f u
d e r f u
d e r f u l
e r f u l #
e r f u l #
r f u l # #
r f u l # #
f u l # # #
How to solve it? Please guide me. Thanks

Answer 1

Add gsub(/ /,"") to the top of #fedorqui's answer to your previous question, change ## to ### and change 5 to 7 and you get:
$ cat tst.awk
{
gsub(/ /,"")
n=length($0)
$0 = "###" $0 "###"
gsub(/./, "& ")
for (i=1; i<=2*n; i+=2)
print substr($0, i, 7*2-1)
print ""
}
$ awk -f tst.awk file
# # # h e l l
# # h e l l o
# h e l l o #
h e l l o # #
e l l o # # #
# # # w o n d
# # w o n d e
# w o n d e r
w o n d e r f
o n d e r f u
n d e r f u l
d e r f u l #
e r f u l # #
r f u l # # #