CF883C Downloading B++

Only $ T $ milliseconds left before the start of well-known online programming contest Codehorses Round 2017. Polycarp needs to download B++ compiler to take part in the contest. The size of the file is $ f $ bytes. Polycarp's internet tariff allows to download data at the rate of one byte per $ t_{0} $ milliseconds. This tariff is already prepaid, and its use does not incur any expense for Polycarp. In addition, the Internet service provider offers two additional packages: - download $ a_{1} $ bytes at the rate of one byte per $ t_{1} $ milliseconds, paying $ p_{1} $ burles for the package; - download $ a_{2} $ bytes at the rate of one byte per $ t_{2} $ milliseconds, paying $ p_{2} $ burles for the package. Polycarp can buy any package many times. When buying a package, its price ( $ p_{1} $ or $ p_{2} $ ) is prepaid before usage. Once a package is bought it replaces the regular tariff until package data limit is completely used. After a package is consumed Polycarp can immediately buy a new package or switch to the regular tariff without loosing any time. While a package is in use Polycarp can't buy another package or switch back to the regular internet tariff. Find the minimum amount of money Polycarp has to spend to download an $ f $ bytes file no more than in $ T $ milliseconds. Note that because of technical reasons Polycarp can download only integer number of bytes using regular tariff and both packages. I.e. in each of three downloading modes the number of downloaded bytes will be integer. It means that Polycarp can't download a byte partially using the regular tariff or/and both packages.

The first line contains three integer numbers $ f $ , $ T $ and $ t_{0} $ ( $ 1

Print the minimum amount of money that Polycarp needs to pay to download B++ compiler no more than in $ T $ milliseconds. If there is no solution, print the only integer -1.

In the first example Polycarp has to buy the first additional package 5 times and do not buy the second additional package. He downloads 120 bytes (of total $ 26·5=130 $ bytes) in $ 120·8=960 $ milliseconds ( $ 960