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Patriots RB Dion Lewis named AFC Offensive Player of the Week after career-day against the Bills

For the second straight week, the conference's top offensive player comes from the Patriots.

Buffalo Bills v New England Patriots Photo by Adam Glanzman/Getty Images

For the second straight week, the AFC Offensive Player of the Week is a member of the New England Patriots: After tight end Rob Gronkowski won the award last week due to his dominating game against the Pittsburgh Steelers, running back Dion Lewis received the honor today following a career performance versus the Buffalo Bills on Sunday.

Lewis set new career-highs in multiple statistical categories during New England's 37-16 victory, including rushing attempts (24) and rushing yards (129). The 27-year old also scored twice and recorded a team-high five catches, gaining 24 yard – all while also serving as the Patriots' kick returner and running back two kickoffs for a combined 43 yards.

Lewis being named AFC Offensive Player of the Week marks the seventh time this season that a member of the Patriots receives the honor: Quarterback Tom Brady was the conference's offensive player of the week in weeks two, three and 10. Gronkowski, as noted above, earned it in week 15. Stephen Gostkowski was named the AFC's top special teams player in week 11 while Lewis himself received the honor one week before that.