According to reports, the Los Angeles and superstar center fielder Mike Trout have agreed to the largest contract in sports history.

The record-shattering deal is for 12 years and an incredible $426.5 million, just under $100 million more than the contract signed by Bryce Harper just two weeks ago. The mega-deal will keep Trout in southern California through his age-38 season. It also breaks the record for Average Annual Value (AAV) at about $35.5 million per season, edging Zack Greinke’s $34.4 million AAV. The deal reportedly has no opt-outs, similar to Harper’s contract, and will likely include a no-trade clause as well. The deal will absorb the final two years of a 6-year $144.5 million contract Trout signed in 2014.

Trout is widely recognized as the best player in baseball and has finished no lower than fourth in the Most Valuable Player (MVP) voting in any of his seven full big league seasons, winning the award in 2014 and 2016.

The 27-year-old is coming off one of his best seasons, finishing with a .312 batting average, 39 home runs and an OPS (On-base + Slugging Percentage) of 1.088 in 140 games.

Trout’s Wins Above Replacement (WAR) numbers are at a historic pace. WAR is a way to measure a player’s overall value to a team. With a 64.3 career WAR, Trout is on pace to set the record for the highest WAR in baseball history.

Trout is a face of Major League Baseball and this deal brings an end to the 2020 free agency frenzy before it even began. New Philadelphia Phillies slugger Bryce Harper was vocal about trying to recruit Trout to Philly, which is about an hour drive north of Trout’s hometown of Millville, New Jersey. The deal puts an end to that discussion.

It had been debated for the last few years as to just how much Trout would command in a new contract, but it seems $426.5 million is the number. Trout and the Angels begin their season next Thursday, March 28, when they head to Oakland to take on the Athletics.

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