linux command to fetch path of S3

I have S3 bucket path like

s3://my-dev-s3/apyong/output/public/file3.gz.tar

I want to fetch all characters after first "/" (not //) and before last "/" .
So here output should be

apyong/output/public

I tried awk -F'/' '{print $NF}' .But its not producing correct results.

>Solution :

Here is sed solution:

s='s3://my-dev-s3/apyong/output/public/file3.gz.tar'
sed -E 's~.*//[^/]*/|/[^/]*$~~g' <<< "$s"

apyong/output/public

Pure bash solution:

s='s3://my-dev-s3/apyong/output/public/file3.gz.tar'
r="${s#*//*/}"
echo "${r%/*}"

apyong/output/public