OXNARD, Calif. — With Ezekiel Elliott‘s holdout at four days, the Dallas Cowboys will sign veteran Alfred Morris, the team announced Monday night.

Sources confirmed a report by 105.3 The Fan in Dallas that Elliott was flying to Cabo as the Cowboys held their first padded practice of camp on Tuesday. While on suspension in 2017, Elliott spent time in Cabo. The Cowboys continue to speak with his agent regarding a contract extension, and they have swapped proposals, but sources indicate that nothing is close at the moment.

Morris played for the Cowboys in 2016-17 and was with the San Francisco 49ers last season. In his two seasons with Dallas, Morris had 790 yards and three touchdowns on 184 yards. He took over as the starter in Elliott’s absence in 2017, including a 127-rushing yard effort.

With Elliott missing, the Cowboys entered camp with just one tailback with a carry in a regular-season game. Darius Jackson, who has taken most of the snaps with the first team in the first three days of practice, had six carries for 16 yards in last year’s season finale. The Cowboys selected Tony Pollard in the fourth round and Mike Weber in the seventh round of the recent draft, and they have Jordan Chunn, who spent last season on the practice squad.

With Elliott on the reserve/did not report list, the Cowboys do not have to make a roster move to add Morris.

Elliott is subject to a $40,000 fine for each day he misses. Once Elliott misses six days of camp, the Cowboys can come after 15% of the prorated portion of the signing bonus he received in 2016, which would be roughly $613,000. For each day after that, they can add 1% up to a max of 25%. If he does not report by Aug. 6, he will not earn an accrued season toward free agency.

Elliott is under contract through 2020, with the Cowboys picking up his $9.09 million fifth-year option. He is set to make $3.853 million.