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Brian Windhorst: Collin Sexton Is Only "Untouchable" Player On Cavs Roster

A general view of the Cleveland Cavaliers home court as fire is sprayed into the air.

CLEVELAND, OH - MAY 7: A general view of Quicken Loans Arena prior to the game between the Cleveland Cavaliers and the Toronto Raptors for Game 4 of the second round of the Eastern Conference playoffs at on May 7, 2018 in Cleveland, Ohio. (Photo by Jason Miller/Getty Images)

NBA free agency opened with a bang earlier this week when LeBron James announced he is headed to Los Angeles. As LeBron looks to become the next great Laker, Cleveland is left wondering how to move on from the King.

Of course, the team, and the city, have experience in moving on from LeBron when he left for Miami to join Dwyane Wade and Chris Bosh. However, this time is a little different, considering he brought a championship to the city.

In the wake of LeBron's exit, the Cavaliers will have to pick up the pieces and move on.

That's easier said than done, especially when the best player on your current roster is Kevin Love. We've not holding anything against Love, but it takes more than one All-Star to deliver a championship, as evidenced by the Warriors and Cavs.

Heading into the 2018-19 season, the Cavs reportedly only have one "untouchable" player currently on the roster.

Collin Sexton.

Brian Windhorst reported that the No. 8 overall pick is the only piece the Cavaliers have made untouchable by other teams.

It will be interesting to see what the Cavaliers look like without LeBron. In the four seasons he was in Miami, Cleveland had the worst winning percentage over that time span.

Will it happen again?

Stay tuned for future free agency madness.