CF173A Rock-Paper-Scissors

Nikephoros and Polycarpus play rock-paper-scissors. The loser gets pinched (not too severely!). Let us remind you the rules of this game. Rock-paper-scissors is played by two players. In each round the players choose one of three items independently from each other. They show the items with their hands: a rock, scissors or paper. The winner is determined by the following rules: the rock beats the scissors, the scissors beat the paper and the paper beats the rock. If the players choose the same item, the round finishes with a draw. Nikephoros and Polycarpus have played $ n $ rounds. In each round the winner gave the loser a friendly pinch and the loser ended up with a fresh and new red spot on his body. If the round finished in a draw, the players did nothing and just played on. Nikephoros turned out to have worked out the following strategy: before the game began, he chose some sequence of items $ A=(a_{1},a_{2},...,a_{m}) $ , and then he cyclically showed the items from this sequence, starting from the first one. Cyclically means that Nikephoros shows signs in the following order: $ a_{1} $ , $ a_{2} $ , $ ... $ , $ a_{m} $ , $ a_{1} $ , $ a_{2} $ , $ ... $ , $ a_{m} $ , $ a_{1} $ , $ ... $ and so on. Polycarpus had a similar strategy, only he had his own sequence of items $ B=(b_{1},b_{2},...,b_{k}) $ . Determine the number of red spots on both players after they've played $ n $ rounds of the game. You can consider that when the game began, the boys had no red spots on them.

The first line contains integer $ n $ ( $ 1

Print two space-separated integers: the numbers of red spots Nikephoros and Polycarpus have.

In the first sample the game went like this: - R - R. Draw. - P - S. Nikephoros loses. - S - P. Polycarpus loses. - R - P. Nikephoros loses. - P - R. Polycarpus loses. - S - S. Draw. - R - P. Nikephoros loses. Thus, in total Nikephoros has $ 3 $ losses (and $ 3 $ red spots), and Polycarpus only has $ 2 $ .