After revealing that he wouldn’t be returning to the Tampa Bay Buccaneers, linebacker Kwon Alexander has already found a new team, signing a deal with the San Francisco 49ers earlier today.

The deal, according to NFL Media’s Ian Rapoport, will be four years in length and worth $54 million, with $27 million guaranteed according to ESPN’s Adam Schefter.

Alexander is coming off of a 2018 season that ended after just six games, when the 24-year old linebacker tore his ACL.

In the games he did play, Alexander was a force to be reckoned with, totaling 45 tackles, a sack, and two forced fumbles for the Bucs. Originally selected by the Buccaneers in the fourth round (with the 124th overall pick) of the 2015 NFL Draft, Alexander has steadily improved his play throughout his career, culminating in a Pro Bowl appearance in 2017.

Despite the health issues, Alexander has proven to be one of the better defensive players available in this years free agency class, and he’ll now head to a 49ers defense that could use some help getting to the quarterback.

Initial reports indicated that he would be looking for around $12-13 million annually, and he’ll get just north of that $13 million at his current deal.

Should he able to avoid any injuries, he’ll help bolster a 49ers team that looks to get right back into contention with the return of quarterback Jimmy Garoppolo.

Last year, the 49ers limped to a 4-12 record thanks in part to the sudden injury to their starting quarterback, but will now find themselves with a very high draft pick, a healthy quarterback, and an influx of new talent.