March 15 (UPI) -- Five-time Pro Bowl selection Rob Gronkowski and star pass-rusher Shaquil Barrett are returning to the Tampa Bay Buccaneers for the 2021 season.
Gronkowski agreed to a one-year deal worth $10 million, his agent, Drew Rosenhaus, told ESPN and NFL Media on Monday. According to the outlets, the contract is for $8 million and can reach $10 million due to incentives.
Earlier Monday, Rosenhaus confirmed that Barrett agreed to a four-year contract worth up to $72 million with the Buccaneers. His deal includes $36 million fully guaranteed.
Gronkowski and Barrett were scheduled to become free agents at the start of the NFL's new league year Wednesday. Instead, the Bucs continued to keep more of their Super Bowl LV team intact entering the 2021 season.
The 31-year-old Gronkowski finished the regular season with 45 catches for 623 yards and seven touchdowns. However, the four-time Super Bowl champion peaked at the right time, as he caught two touchdowns in the Buccaneers' 31-9 win over the Kansas City Chiefs in Super Bowl LV.
Barrett, a Pro Bowl selection in 2019, recorded 57 total tackles, eight sacks, two forced fumbles and three passes defensed last season.
