Di two-year-old daughter of one Tampa Bay Buccaneers NFL player don drown for one swimming pool for di family house, authorities confam.
Police officers bin respond to one call say Arrayah, di youngest daughter of Shaquil Barrett, bin fall inside di pool around 09:30 local time (14:30 BST).
Dem carry her go hospital for Tampa, Florida, wia dem pronounce her dead.
Di Super Bowl winning linebacker wit di Buccaneers, Barrett, wey be 30 years, and im wife get three oda children.
Di Buccaneers don release statement on di "tragic" and "heart-breaking" news
"Our thoughts and prayers dey wit Shaq, (im wife) Jordanna and di entire Barrett family during dis unimaginably difficult time," e tok.
"While no words fit provide true comfort for dis kain time, we offer our support and love as dey begin to process dis very profound loss of dia beloved Arrayah."
Police say dem don begin investigation but dem no get suspicion about di death.
Na last week Arrayah bin turn 2 years, and her father celebrate her on im Instagram account.
“Happy 2nd bday to mu cutie girl. So sweet and cute. To made our family complete. I love you baby girl.” Barrett write.
Who be Shaquil Barrett?
Oga Barrett dey enta im fifth year wit di Tampa Bay Buccaneers, afta e spend di first four seasons of im career wit di Denver Broncos.
E currently dey recover from one Achilles injury wey don keep am for bench during di second half of last season.
For 2019, e bin lead di NFL wit 19.5 sacks and for di following season, e help di Buccaneers win di Super Bowl alongside di retired, popular NFL quarterback Tom Brady.
Barrett and wife Jordanna be high school sweethearts.
Dem born Arrayah two months afta Barrett and di Bucs beat Kansas City Chiefs for Super Bowl 55. E bin just sign one four-year, $72 million contract wit $34 million guaranteed.