CF989E A Trance of Nightfall

The cool breeze blows gently, the flowing water ripples steadily."Flowing and passing like this, the water isn't gone ultimately; Waxing and waning like that, the moon doesn't shrink or grow eventually." "Everything is transient in a way and perennial in another." Kanno doesn't seem to make much sense out of Mino's isolated words, but maybe it's time that they enjoy the gentle breeze and the night sky — the inexhaustible gifts from nature. Gazing into the sky of stars, Kanno indulges in a night's tranquil dreams. There is a set $ S $ of $ n $ points on a coordinate plane. Kanno starts from a point $ P $ that can be chosen on the plane. $ P $ is not added to $ S $ if it doesn't belong to $ S $ . Then the following sequence of operations (altogether called a move) is repeated several times, in the given order: 1. Choose a line $ l $ such that it passes through at least two elements in $ S $ and passes through Kanno's current position. If there are multiple such lines, one is chosen equiprobably. 2. Move to one of the points that belong to $ S $ and lie on $ l $ . The destination is chosen equiprobably among all possible ones, including Kanno's current position (if it does belong to $ S $ ). There are $ q $ queries each consisting of two integers $ (t_i, m_i) $ . For each query, you're to help Kanno maximize the probability of the stopping position being the $ t_i $ -th element in $ S $ after $ m_i $ moves with a proper selection of $ P $ , and output this maximum probability. Note that according to rule 1, $ P $ should belong to at least one line that passes through at least two points from $ S $ .

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The points in $ S $ and possible candidates for line $ l $ are depicted in the following figure. For the first query, when $ P = (-1, -3) $ , $ l $ is uniquely determined to be $ 3x = y $ , and thus Kanno will move to $ (0, 0) $ with a probability of $ \frac 1 2 $ . For the third query, when $ P = (2, 2) $ , $ l $ is chosen equiprobably between $ x + y = 4 $ and $ x = y $ . Kanno will then move to the other four points with a probability of $ \frac 1 2 \cdot \frac 1 3 = \frac 1 6 $ each, or stay at $ (2, 2) $ with a probability of $ \frac 1 3 $ .