The Super Bowl champion Tampa Bay Buccaneers keep on getting the band back together.

The most recent to rejoin: Rob Gronkowski.

The star tight end and the Bucs have consented to a 1-year bargain worth $10 million, his representative, Drew Rosenhaus, disclosed to ESPN’s Adam Schefter on Monday.

Gronkowski was one of a few high-profile Bucs players – defensive lineman Ndamukong Suh, wide receiver Antonio Brown, kicker Ryan Succop and running back Leonard Fournette were among the others – whom the team had wanted to bring back after a month ago’s Super Bowl win.

Prior Monday, the Bucs and star pass-rusher Shaquil Barrett consented to a four-year deal worth up to $72 million, Rosenhaus told Schefter.

The Buccaneers exchanged a fourth-round draft pick to the New England Patriots last offseason for Gronkowski’s privileges and a seventh-round pick. Gronkowski, who had resigned from the NFL in 2019, got going the season gradually, shaking off some rust in the wake of being away from football for a year and finding a part in Bruce Arians’ offense, which wasn’t known for being tight end friendly.

While friendly organizer Byron Leftwich and quarterback Tom Brady attempted to cut out a role for Gronkowski, he accepted his role as a blocker and as a leader and completed the regular season with 45 receptions for 623 yards and 7 touchdowns.

He peaked at the right time, however, getting done with two touchdowns in the Buccaneers’ 31-9 triumph in Super Bowl LV as he and Brady broke Joe Montana’s and Jerry Rice’s postseason record for most touchdowns by a QB-receiving duo.

Gronkowski, 31, spent the initial nine seasons of his NFL vocation with the Patriots, winning three Super Bowls with the team. He was a five-time Pro Bowl selection and four-time first-team All-Pro.

He has 566 receptions for 8,484 yards and 86 touchdowns in 131 regular-season games in his career.