2006 AIME II Problem 11

A sequence is defined as follows a_1=a_2=a_3=1, and, for all positive integers n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. Given that a_{28}=6090307, a_{29}=11201821, and a_{30}=20603361, find the remainder when \sum^{28}_{k=1} a_k is divided by 1000.

Leading zeroes must be inputted, so if your answer is 34, then input 034

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.