Bonus Cards

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Problem

Dmitry loves programming competitions. This year the Champions League Finals are held in the city where Dmitry lives, so he wants to watch the games from the stands. The competition is popular, and most tickets are already set aside for VIPs and for sponsors.

Tickets that reach the general public are handed out like this. Anyone who wants to watch submits a request stating which payment method they will use for the ticket. Suppose nn seats are left. The draw runs in several rounds. In each round, every request that has not won yet receives a number of draw slots decided by its payment method. One slot is picked uniformly at random out of all slots in that round, and the request holding it wins. A request that has won drops out of the following rounds. The draw finishes after round nn, or earlier if no request is left without a ticket. International Card Processing Corporation (ICPC) sponsors the Champions League, so a request that states an ICPC card gets two draw slots in each round, and a request that states any other payment method gets one.

Dmitry has an ICPC card, and he also has an Advanced Credit Merchandise (ACM) card that pays a bonus back on everything he spends. His brother Petr works at the company that runs the draw, so he told Dmitry in advance how many people had asked with an ICPC card and how many had asked with another payment method. Before choosing a card, Dmitry wants to know his probability of getting a ticket with the ICPC card and with the ACM card. Dmitry's request is added on top of the requests Petr told him about.

Find both probabilities.

Input

The input is a single line with three integers: the number of seats in the draw nn (1n30001 \le n \le 3000), the number of requests stating an ICPC card aa, and the number of requests stating another payment method bb (0a,b1090 \le a, b \le 10^9).

Output

Print the probability that Dmitry gets a ticket with the ICPC card on the first line, and the probability that he gets a ticket with the ACM card on the second line. Print both probabilities rounded to exactly six digits after the decimal point. In every input the two probabilities sit well away from a rounding boundary.