Gronkowski signs six-year deal with Pats
Rob Gronkowski agreed to stay with the New England Patriots for another six years after signing an extended contract through 2019, according to a report on the NFL's website.
Although the Patriots did not announce the deal, the report on the National Football League's (NFL) website said Gronkowski agreed to a $54 million contract.
The new deal, which was tagged on to his rookie contract, would make him the highest paid tight end in the history of NFL.
According to sources cited by NFL.com, Gronkowski's deal included an $8 million signing bonus, $13 million guaranteed for skill and $18 million guaranteed for injury.
Gronkowski was a second-round pick in the 2010 NFL Draft and set single-season records for tight ends last season with 1,327 receiving yards and 17 touchdowns on 90 receptions.
But he suffered an ankle injury in the AFC Championship Game win over Baltimore and was restricted in his movement during the Super Bowl loss to the New York Giants.