Find the least positive four-digit solution $r$ of the congruence $r^2 + 4r + 4 \equiv r^2 + 2r + 1 \pmod{55} $.
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Find the least positive four-digit solution $r$ of the congruence $r^2 + 4r + 4 \equiv r^2 + 2r + 1 \pmod{55} $.
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
1016
Find the least positive four-digit solution $r$ of the congruence $r^2 + 4r + 4 \equiv r^2 + 2r + 1 \pmod{55} $.
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