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Giants agree to deal with free-agent DT A'Shawn Robinson

The Giants are adding more experience to their defensive front.

Veteran defensive tackle A'Shawn Robinson has agreed to a one-year deal with New York, NFL Network Insiders Ian Rapoport and Mike Garafolo reported on Monday. Robinson's contract is worth up to $8 million, per Rapoport and Garafolo. The team has since announced the signing.

Robinson's move to New York will pair him with his third team in his eight-year NFL career. The defensive tackle entered the league as a second-round pick of the Lions out of Alabama in 2016, spending four seasons with the Lions before moving west to join the Rams in 2020. He won a Super Bowl with the Rams in the 2021 season before appearing in 10 games in his final campaign with Los Angeles in 2022.

At 28 years old, Robinson goes from one contender to a team aiming to take the next step toward Super Bowl contention following a surprising 2022 season in which the Giants reached the postseason as a wild-card qualifier and upset the Vikings at home during Super Wild Card Weekend before bowing out in a blowout loss to the eventual NFC champion Eagles.

Robinson joins a defense that has welcomed a number of additions this offseason, including linebacker Bobby Okereke, cornerback Bobby McCain and defensive tackle Rakeem Nunez-Roches. Robinson slots in alongside Nunez-Roches as a seasoned defender who can form a quality rotational group behind starters Leonard Williams and Dexter Lawrence with hopes to improve from the Giants' 25th-place finish in total defense from a year ago.

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