Ezekiel Elliott has found a new home. The former Ohio State standout is reportedly signing with the New England Patriots.

According to Ian Rapoport and Tom Pelissero, Elliott is signing a 1-year deal worth up to $6 million.

After a decorated Ohio State career that included MVP honors in the College Football Playoff National Championship, Elliott was drafted No. 4 overall by the Dallas Cowboys in the 2016 NFL Draft. He spent his first 7 professional seasons in Dallas.

Elliott’s career got off to a strong start. He picked up PFWA All-Rookie Team, first-team All-Pro and Pro Bowl recognition for his standout rookie season after leading the NFL in rushing yards with 1,631, his most productive season as a ball carrier.

In 2018, Elliott led the NFL in rushing again, logging 1,434 yards on the ground and a career-high 567 receiving yards. He was named second-team All-Pro honors and selected to his second Pro Bowl.

Ahead of the 2019 season, Elliott held out of camp for a new contract. He signed a 6-year deal worth up to $90 million with $50 million guaranteed. Though the deal went until 2026, he was released by the Cowboys on March 15.

Over his career, Elliott has played in 103 regular-season games, rushing 1,881 times for 8,262 yards and 68 touchdowns, along with 305 catches for 2,336 yards and 12 scores. He’s coming off a 2022 season in which he rushed 231 times for 876 yards, catching 17 passes for 92 yards.