A pair of Bengals wide receivers had big games in Week 8, but only one of them could be named the AFC offensive player of the week.

The NFL announced that Tee Higgins was their choice for that award on Wednesday. Higgins had 10 catches for 155 yards and two touchdowns, which gave him the edge over Ja’Marr Chase. Chase finished the 43-36 overtime win with 11 catches for 141 yards and a touchdown.

Higgins’s first score gave the Bengals a 21-13 lead in the third quarter and his second tied the game with 1:17 left to play in regulation.

It is the first time in his career that Higgins has taken weekly honors in the conference.