While Conor McGregor seems to be on a collision course with the winner of Saturday’s UFC 179 featherweight title bout between Jose Aldo and Chad Mendes, he must first get past Dennis Siver at UFC Fight Night 59.

McGregor (16-2 MMA, 4-0 UFC) doesn’t see that being much of an issue, though, as he believes Siver (22-9 MMA, 11-6 UFC) is nothing more than another stepping-stone toward a championship fight.

“Dennis is an easy contest,” McGregor said today at a UFC 179 Q&A in Brazil. “I like to think of Dennis as a keep fresh, get rich fight.”

UFC President Dana White previously said McGregor could face Aldo or Mendes next, but “The Notorious” later declared he’s eager to keep active and doesn’t want to wait for a title shot.

He got his wish in a matchup first reported by MMAjunkie, as the UFC today confirmed McGregor is set to meet Siver in the UFC Fight Night 59 headliner on Jan. 18 at Boston’s TD Garden.

Even though Aldo (24-1 MMA, 6-0 UFC) and Mendes (16-1 MMA, 7-1 UFC) are on schedule to collide this weekend, McGregor is currently the center of attention in Rio de Janeiro. The brash Irishman hosted a fan Q&A prior to the UFC 179 weigh-ins and was bombarded with jeers and taunts by the Brazilian fans.

McGregor didn’t back down from any of it, though, as he stood firm with the notion he’s the No. 1 contender in the featherweight division. The 26-year-old said he believes Aldo will defend against Mendes at UFC 179, and after he takes care of business against Siver in January, he looks forward challenging “Scarface” for UFC gold.

“I am the No. 1 contender,” McGregor said. “The next time Jose steps in the octagon after Saturday night will be to face me. In the meantime, I took a fight with Dennis Siver to eliminate another contender. It’s as simple as that.”

UFC 179 goes down from Rio de Janeiro’s Maracanazinho Gymnasium on Saturday. Aldo vs. Mendes headlines the pay-per-view main card following prelims on FOX Sports 1 and UFC Fight Pass.

For more on UFC Fight Night 59, check out the UFC Rumors section of the site.