AT_abc407_b [ABC407B] P(X or Y)

Two dice, each with six faces $ 1,2,3,4,5,6 $ , are rolled. Find the probability that at least one of the following two conditions holds: - The sum of the two outcomes is at least $ X $ . - The absolute difference of the two outcomes is at least $ Y $ . Each face of each die is equally likely, and the two dice are independent.

The input is given from Standard Input in the following format: > $ X $ $ Y $

Output the probability that the two outcomes satisfy at least one of the two conditions. Your answer is accepted if its absolute error from the true value is at most $ 10^{-9} $ .

### Sample Explanation 1 Let $ (x,y) $ denote the event that the dice show $ x $ and $ y $ . - The sum is at least $ 9 $ for $ (3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6) $ . - The difference is at least $ 3 $ for $ (1,4),(1,5),(1,6),(2,5),(2,6),(3,6),(4,1),(5,1),(5,2),(6,1),(6,2),(6,3) $ . At least one of these conditions holds for the following $ 20 $ pairs: $ (1,4),(1,5),(1,6),(2,5),(2,6),(3,6),(4,1),(4,5),(4,6),(5,1),(5,2),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) $ . Thus, the answer is $ \dfrac{20}{36}=\dfrac59=0.5555555555\ldots $ . Because an absolute error of at most $ 10^{-9} $ is allowed, outputs such as `0.5555555565` or `0.55555555456789` are accepted. ### Sample Explanation 2 Neither the sum of two dice can be $ 13 $ or greater, nor can their difference be $ 6 $ or greater. Thus, the answer is $ 0 $ . ### Constraints - $ 2 \le X \le 13 $ - $ 0 \le Y \le 6 $ - All input values are integers.