NEW ORLEANS (WGNO) — After an outstanding performance against Seattle over the weekend, New Orleans Saints tight end Taysom Hill has been named NFC Offensive Player of the week, the NFL announced on Wednesday (Oct. 12).

In Sunday’s matchup against the Seahawks, Hill was instrumental in snapping a three-game losing streak for the Saints. Although listed as a tight end on the Saints’ roster, Hill confirmed he can be versatile after rushing three TDs, including a 60-yard run in the fourth quarter that would mark a career-high 112 rushing yards.

The Saints went on to defeat the Seahawks 39-32.

In a post-game interview, Hill said that the game was one of his best performances in the NFL.

“I think my mindset is, man I’m going to take advantage of every opportunity, and I felt like I was going to get more opportunity this week and so I did my part to make sure I was prepared for when it came up,” the tight end told media after the game on Sunday. “I feel like we’ve kind of been at this tipping point and this was the game that we were ready for.”

Hill’s performance on Sunday marks him as the second Saints tight end to clutch the honor after Jimmy Graham in 2012 and 2013. He is the third player ever to record at least 100 rushing touchdowns and a TD pass in a single game, succeeding Ronnie Brown and Pro Football Hall of Famer LaDainian Tomlinson. This is Hill’s second time earning Player of the Week honors.

Other players receiving NFC honors this week were Dallas Cowboys linebacker Micah Parsons and Philadelphia Eagles kicker Cameron Dicker with Defensive and Special Teams Players of the Week, respectively.

This weekend, Hill and the Saints will take on the Joe Burrow-Ja’Marr Chase duo with a home game against the Bengals. That game kicks off at 12 p.m.