<image>As shown in the figure, ray $$OA$$ coincides with $$OM$$. Taking $$O$$ as the rotation center, rotate $$OA$$ counterclockwise: $$OA ⇒ O{A}_{1} ⇒ O{A}_{2} ⇒ …⇒ O{A}_{n} …$$. The rotation angles are $$∠AO{A}_{1}={2}^{\circ}, ∠{A}_{1}O{A}_{2}={4}^{\circ}, ∠{A}_{2}O{A}_{3}={8}^{\circ}$$, and so on. The next rotation angle (not exceeding $${360}^{\circ}$$) is twice the previous rotation angle. When the rotation angle exceeds $${360}^{\circ}$$, it restarts from $${2}^{\circ}$$, i.e., $$∠{A}_{8}O{A}_{9}={2}^{\circ}, ∠{A}_{9}O{A}_{10}={4}^{\circ}$$, and so on cyclically.

Find the smallest value of $$n$$ such that $$O{A}_{n}$$ coincides with $$ON$$. 

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