One of the key pieces of the Cleveland Indians' explosive offense in 2017 is apparently heading back to the Big Apple.

Wednesday night, ESPN's Jerry Crasnick was the first to report that free agent outfielder Jay Bruce has agreed to a three-year, $39 million contract to rejoin the New York Mets.

Bruce was a surprising pickup by the Tribe on August 9, 2017, acquiring him from the Mets in a non-waiver trade. The Indians picked up approximately $5 million dollars of Bruce's 2016 salary to make the trade possible.

After the trade, Bruce hit .248 with 7 HR and 26 RBI in 43 regular season games. He added a pair of homers during the Tribe's American League Division Series loss to the Yankees. For the season, Bruce hit a career-high 36 homers and drove in 101 runs.

Bruce's acquisition helped to fuel the Indians' American League-record 22-game winning streak in late August and into September. His walk-off hit in extra innings against the Royals on September 14 shattered the record.

The fact that Bruce wasn't signed by a team earlier or for more money was somewhat of a surprise. According to The Athletic/Fox Sports' Ken Rosenthal, the three-time All-Star's deal has been backloaded with more money in the final two years of his deal.

Bruce has never played a full season for the Mets. He was acquired from Cincinnati in 2016 and played 50 games for New York that season, followed by 103 games in 2017 before coming to the Tribe.