DAYTON, Ohio (AP) — No. 11 seed St. Bonaventure (25-7) vs. No. 11 seed UCLA (21-11)
First Four, East region; Dayton, Ohio; 9:10 p.m. EDT.
BOTTOM LINE: UCLA makes its 49th tournament appearance but its debut in the First Four. The Bruins lost to Kentucky in a regional semi-final last season. St. Bonaventure won Atlantic 10 regular season and back are back in the field for the first time since 2012 and seventh all time. Both are looking for their first NCAA Tournament victories. The winner gets No. 6 seed Florida on Thursday.
BONNIES BELONG: St. Bonaventure is driven by the guard play of seniors Jaylen Adams (19.8 points per game) and Matt Mobley (18.5). The Atlantic 10 regular-season winner has victories this season over Maryland, Syracuse and Rhode Island.
UCLA IN DAYTON?: Coach Steve Alford, who has led the Bruins to the tournament for the fourth time in five seasons, said he was surprised they were picked last and were sent to Dayton. He thought UCLA’s resume — wins over Kentucky and Arizona, and a two-game sweep of USC — warranted a better seeding. After all, the Bruins have been to 18 Final Fours.
LIGHTING UP THE SCOREBOARD: The Bruins are second in the Pac-12 in scoring (81.9 points per game) and lead the conference in rebounding. Junior guard Aaron Holiday is an All-Pac-12 selection who averages a conference-best 20.3 points per game.
DID YOU KNOW: Florida, which awaits the winner of the play-in game, has beaten UCLA three times in the NCAA Tournament since 2006. The most recent was a regional semifinal in 2014, Alford’s first season.
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