Former Raiders TE Lee Smith signs 3-year deal with Bills

After a meeting with Buffalo, and plenty of competition from other teams, former Raiders Tight End Lee Smith has chosen to sign with the Bills to continue his NFL career, according to NFL Insider Adam Caplan.

#Bills TE Lee Smith's 3-year deal is worth $9m with $4.9m fully gtd–2-year structure (year 1: $1.9m base, $1.5m 5/15 RB; year 2: $1m of base, $500k RB). They had a lot of competition for him, which might help to explain the strong guarantees.

— Adam Caplan (@caplannfl) May 23, 2019

Smith originally played for Buffalo from 2011-2015, totaling just 20 catches for 144 yards. After his tenure in New York, Lee decided to sign with Oakland for a 3-year, $9 million contract, coincidentally the same contract he will be signing with the Bills,

In 2018, Smith had the best year of his career, finishing with 10 catches for 73 yards, however, three of the catches went for touchdowns.

With the Raiders already stock with Tight Ends for the upcoming season, it was not a shock to see the team let a player find a new home. It will be interesting to see whether Smith can find the end-zone with second-year quarterback Josh Allen.