The Baltimore Orioles agreed to terms with free agent outfielder Ramon Laureano on a one-year contract with a club option for 2026, the club announced Tuesday afternoon.

Terms were not disclosed, but multiple reports had the deal worth $4 million.

Laureano, 30, played for the Cleveland Guardians and Atlanta Braves last season and batted a combined .259 with a .311 on-base percentage and .437 slugging percentage with 11 home runs and 33 RBIs in 98 games.

He played better for the Braves, batting .296 with 10 homers and 29 RBIs in 67 games — with 16 multihit games, including five three-hit games.

Laureano is a career .247 hitter with 82 homers, 252 RBIs, a .319 on-base percentage and .430 slugging percentage for the Oakland Athletics (2018-23), Guardians (2023-24) and Braves.

The Houston Astros selected the Dominican Republic native in the 16th round of the 2014 MLB Draft out of Northeastern Oklahoma A&M. He was traded from Houston to Oakland in November 2017 and selected off waivers by the Guardians in August 2023. The Braves signed him as a free agent on May 29, 2024, four days after the Guardians released him.

The O’s opened a spot on the 40-man roster for Laureano by designating infielder Luis Vazquez for assignment.