fix POD errors reported by newer pod2man

1 files changed, 17 insertions, 17 deletions

diff --git a/pod/perlretut.pod b/pod/perlretut.pod
index 76522c6288..45d66b65c1 100644
--- a/pod/perlretut.pod
+++ b/pod/perlretut.pod
@@ -642,50 +642,50 @@ of what Perl does when it tries to match the regexp
 
 =over 4
 
-=item 0
+=item Z<>0
 
 Start with the first letter in the string 'a'.
 
-=item 1
+=item Z<>1
 
 Try the first alternative in the first group 'abd'.
 
-=item 2
+=item Z<>2
 
 Match 'a' followed by 'b'. So far so good.
 
-=item 3
+=item Z<>3
 
 'd' in the regexp doesn't match 'c' in the string - a dead
 end.  So backtrack two characters and pick the second alternative in
 the first group 'abc'.
 
-=item 4
+=item Z<>4
 
 Match 'a' followed by 'b' followed by 'c'.  We are on a roll
 and have satisfied the first group. Set $1 to 'abc'.
 
-=item 5
+=item Z<>5
 
 Move on to the second group and pick the first alternative
 'df'.
 
-=item 6
+=item Z<>6
 
 Match the 'd'.
 
-=item 7
+=item Z<>7
 
 'f' in the regexp doesn't match 'e' in the string, so a dead
 end.  Backtrack one character and pick the second alternative in the
 second group 'd'.
 
-=item 8
+=item Z<>8
 
 'd' matches. The second grouping is satisfied, so set $2 to
 'd'.
 
-=item 9
+=item Z<>9
 
 We are at the end of the regexp, so we are done! We have
 matched 'abcd' out of the string "abcde".
@@ -1255,35 +1255,35 @@ backtracking.  Here is a step-by-step analysis of the example
 
 =over 4
 
-=item 0
+=item Z<>0
 
 Start with the first letter in the string 't'.
 
-=item 1
+=item Z<>1
 
 The first quantifier '.*' starts out by matching the whole
 string 'the cat in the hat'.
 
-=item 2
+=item Z<>2
 
 'a' in the regexp element 'at' doesn't match the end of the
 string.  Backtrack one character.
 
-=item 3
+=item Z<>3
 
 'a' in the regexp element 'at' still doesn't match the last
 letter of the string 't', so backtrack one more character.
 
-=item 4
+=item Z<>4
 
 Now we can match the 'a' and the 't'.
 
-=item 5
+=item Z<>5
 
 Move on to the third element '.*'.  Since we are at the end of
 the string and '.*' can match 0 times, assign it the empty string.
 
-=item 6
+=item Z<>6
 
 We are done!